Ray Optics And Optical Instruments

Given,

Size of the candle, h = 2.5 cm

Object distance, u = –27 cm

Image size = h’ = ?

Radius of curvature of the concave mirror,

R = – 36 cm

Now focal length f is given as f = R/2

∴ f = R/2 = -18 cm

The image distance can be obtained using the mirror equation,

Where,

f is the focal length

u is object distance and

v is the image distance

⇒

⇒

⇒

⇒ v = -54 cm

Therefore, the screen should be placed at a distance of 54 cm

from the mirror to obtain a sharp image.

The magnification is given by

Where,

m is the magnification

h’ is the height of the image

h is the height of the object

v is the image distance

u is the object distance

Using the equation, we get,

⇒

⇒ h^’ = -5 cm

∴ The height of the image is 5cm. The negative sign indicate that the image is real, inverted and magnified.

Given,

Size of the needle, h = 4.5 cm

Object distance, u = – 12 cm

Focal length of the convex mirror, f = 15 cm

The image distance can be obtained using the mirror equation,

Where,

f is the focal length

u is object distance and

v is the image distance

⇒

⇒

⇒

⇒ v = 6.67 cm

Therefore, the image of the needle is 6.67 cm away

from the mirror.

The magnification is given by

Where,

m is the magnification

h’ is the height of the image

h is the height of the object

v is the image distance

u is the object distance

Using the equation, we get,

⇒

⇒ h' = + 2.5 cm

⇒

⇒ m = 0.56

∴ Magnification is 0.56. And, the height of the image is 2.5 cm. The image is virtual.

If the needle is moved farther from the mirror the image will also move farther away. The size of the image will also decrease gradually.

Actual depth of the needle in water, h₁ = 12.5 cm

Apparent depth of the needle in water, h₂ = 9.4 cm

Refractive index of water = μ

The value of μ can be obtained as follows:

According to Snell’s law,

∴ μ = 1.33

Hence, the refractive index of water is about 1.33.

When the water is replaced by a liquid of refractive index,

�^’ = 1.63

The actual depth of the needle remains the same, but the apparent depth of the needle changes.

EXPLANATION: This is because the light ray coming from the needle inside liquid suffers refraction at the liquid-air interface, as the refractive index of liquid is greater than air the light ray bends away from the normal at the point of incidence due to which the refracted ray appears to be coming from a point at a depth less than the actual depth of the needle.

Let h₂’ be the new apparent depth of the needle.

Hence, we can write the relation:

⇒

⇒ h₂’ = 7.67 cm

Hence, the new apparent depth of the needle is 7.67 cm. It is less than h₂.

∴ Distance by which the microscope should move = 9.4 cm - 7.6 cm

= 1.73 cm upwards.

For the glass -air interface:

Angle of incidence, i = 60^°

Angle of refraction, r = 35^°

According to Snell’s law,

⇒

⇒

⇒

Note: refractive index is a unitless quantity.

For the water -air interface:

Angle of incidence, i = 60^°

Angle of refraction, r = 47^°

According to Snell’s law,

⇒

⇒

⇒

⇒

For the glass - water interface:

Angle of incidence, i = 45⁰

Angle of refraction, r = ? ⁰

According to Snell’s law,

⇒

⇒

⇒ r ≈ 37^°

Therefore, the angle of refraction in glass is

Given Data,

Depth of bulb in water = d = 80 cm = 0.8 m

Refractive index of water = μ = 1.33

Construction:

Please consider point “N” at the bottom of the tank where the bulb is placed.

Let the axis line “MOP” be the line on the surface of the water which is 80 cm from the bottom of the tank.

The below is the figure which we would get once we finish our construction.

From the figure we get,

Angle of incidence = Angle of reflection = 90°

Principle: The law of reflection governs the reflection of light-rays off smooth conducting surfaces (here water). The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the water all lie in the same plane. Furthermore, the angle of reflection is equal to the angle of incidence. Both angles are measured with respect to the normal to the water in the tank.

Since we consider bulb as a point source, the emergent light can be considered as a circle.

The radius of this circle can be obtained by,

R = MP/2 (MP is taken as diameter here)

= MO = MP (from the above figure)

According to Snell’s law, ratio of sine of angle of incidence and sine of angle of refraction is always constant for a given pair of media.

μ = sin r/ sin i

⇒ (substituting μ from given data & r from figure)

⇒ i = sin^-1 (1/1.33) = 48.75° (from log tables)

From the figure,

tan i = OP/ON = R/d

⇒ R = d × tan(i)

⇒ R = tan 48.75° x 0.8 = 0.91 m

Area of the surface of water (since we considered the point source is bulb and the surface covered is a circle) = πR² = π(0.91)² = 2.61 m²

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m²

Given Data,

Angle of minimum deviation is δ_m = 40°.

Angle of the prism is 𝐴 = 60°.

Refractive index of water is μ_w = 1.33.

Refractive index of the material of the prism be μ_p .

The angle of deviation is related to refractive index as:

..(1)

Reference Figure for the above formula:

Explanation: There is an angle of incidence at which the sum of the two deflections is a minimum. The deviation angle at this point is called the "minimum deviation" angle, or "angle of minimum deviation". At the minimum deviation angle, the incidence and exit angles of the ray are identical. One of the factors that causes a rainbow is the bunching of light rays at the minimum deviation angle that is close to the rainbow angle.

Substituting the given data in the equation (1):

⇒

⇒

⇒ 𝜇′ = 1.532

Hence, we can conclude that the refractive index of the material of the prism is 1.532.

B. Since the prism is placed in water, let 𝛿′𝑚 be the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

Also,

⇒

⇒

⇒

⇒

⇒

⇒

⇒ (A + 𝛿′𝑚) = 2 × 35.16⁰

⇒ (A + 𝛿′𝑚) = 70.32⁰

⇒ 𝛿′𝑚 = 70.32⁰ – 60⁰(∵ A = 60⁰)

∴ 𝛿′𝑚 = 10.32⁰

Hence the new angle of minimum deviation of a parallel beam of light is 10.32⁰

Given Data:

Refractive index of glass, μ = 1.55

Focal length of the double-convex lens, f = 20 cm

Consider,

Radius of curvature of one face of the lens = R₁

Radius of curvature of the other face of the lens = R₂

Radius of curvature of the double-convex lens = R

Since it is mentioned that both faces are of the same radius,

⇒ R₁ = R and R₂ = -R

The value of R can be calculated as: t

… (1)

(Explanation: The above formula is generated by the ray diagram of the double convex lens. )

Substituting the given data in the equation (1), we get,

⇒ R = 0.55 × 2 × 20

⇒ R = 22 cm

Hence, the radius of curvature of the double-convex lens is 22 cm.

Given data,

The Lens is placed in the path of the convergent beam 12 cm from P.

Here, the object is virtual and the image is real, u = + 12 cm (since, the object on right and virtual)

Reference ray diagram:

a) Given f = + 20 cm for the convex lens.

Using Len’s formula:

⇒

⇒

i.e. v = 7.5 cm (image on right and real)

This means that the beam converges as a point 7.5 cms from the lens

b) Here, f = -16 cm

According to the lens formula, we have the relation:

l

⇒

⇒

⇒ v = 48 cm

Hence, the image will be located 48 cm from the lens.

Given data,

Size of the object, h₁ = 3 cm

Object distance, u = – 14 cm

Focal length of the concave lens, f = – 21 cm

Image distance = v

According to the lens formula, we have the relation:

⇒

⇒

⇒

⇒

⇒

⇒ v = - 8.4 cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

⇒ h₂ = 0.6 × 3 = 1.8 cm

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

Given Data,

Focal length of the convex lens, f₁ = 30 cm

Focal length of the concave lens, f₂ = – 20 cm

Focal length of the system of lenses = f

The equivalent focal length of a system of two lenses in contact is given as:

⇒

⇒

⇒ f = -60 cm

Hence, the focal length of the combination of lenses is 60 cm.

The negative sign indicates that the system of lenses acts as a diverging lens.

Given:

Focal length of the objective lens, f₁ = 2.0 cm

Focal length of the eyepiece lens, f₂ = 6.25 cm

Distance between eyepiece and objective, d = 15 cm

We need to find the distance of object from objective lens for the following two cases:

(a) The final image is formed at least distance of distinct vision, v₂ = 25 cm

Let the distance of object from eyepiece lens = u₂ cm

By using lens maker formula,

…(1)

From equation 1 we have,

…(2)

Where, f₂ = focal length of the eye piece lens

v₂ = Distance of image formation

u₂ = Distance of object from eyepiece

Plugging the values in equation (2)

⇒

⇒ u₂ = -5 cm

We understand that image distance from objective lens v₁, is given by

v₁ = d + u₂

v₁ = 15cm - 5cm

v₁ = 10 cm

Let image distance for the objective lens = u₁

From equation (1), for objective lens, we have,

…(3)

Plugging the values in equation (3) we have,

⇒ u₁ = -2.5 cm

We know that magnifying power of a compound lens is given by,

m = ) …(4)

Where,

m = magnification

v₁ = image distance from objective lens.

u₁ = object distance from objective lens

d^’ = minimum distance of distinct vision (25 cm)

f₂ = Focal length of eyepiece

Now putting the values in equation (4)

m = 20.

Hence the magnification of the compound microscope in this case is 20. Magnification is a unit less quantity, as you can observe from the equation.

(b) Distance of final image formed, v₂ = ∞

Object distance for the eyepiece = u₂

By using lens maker formula we have,

…(1)

From equation 1 we have,

…(2)

Where, f₂ = focal length of the eye piece lens

v₂ = Distance of image formation

u₂ = Distance of object from eyepiece

Plugging the values in equation (2)

We get,

u₂ = -6.25 cm

We understand that image distance from objective lens v₁, is given by

v₁ = d + u₂

v₁ = 15cm – 6.25cm

v₁ = 8.75 cm

Let image distance for the objective lens = u₁

From equation (1), for objective lens, we have,

…(3)

Plugging the values in equation (3) we have,

⇒

⇒ u₁ = -2.59 cm

We know that magnifying power of a compound lens is given by,

m = …(4)

Where,

m = magnification

v₁ = image distance from objective lens.

u₁ = object distance from objective lens

d^’ = minimum distance of distinct vision (25 cm)

f₂ = Focal length of eyepiece

Now putting the values in equation (4)

m = 13.51

Hence, the magnifying power of the microscope in this case is 13.51.

Given:

Focal length of the objective lens, f₀ = 8 mm (0.8 cm)

Focal length of the eyepiece, f_e = 2.5 cm

Object distance for the objective lens, u₀ = -9.0 mm ( -0.9 cm)

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece, v_e = -d = -25 cm

Let object distance for the eyepiece = u_e.

By using lens maker formula,

…(1)

From equation 1 we have,

…(2)

Where, f_e = focal length of the eye piece lens

v_e = Distance of image formation

u_e = Distance of object from eyepiece

Plugging the values in equation (2)

⇒

⇒ u₂ = -2.27 cm

We can find out the distance of image for objective lens using the equation (1)

For objective lens, we have

From equation 1 we have,

…(3)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Plugging the values in equation (3)

⇒ v₀ = 7.2 cm

We know that the distance between the objective lens and eyepiece, d is given by,

d = |u_e| + v_o

d = 2.72 + 7.2

d = 9.47 cm

We know that magnifying power of a compound lens is given by,

m = ) …(4)

Where,

m = magnification

v₀ = image distance from objective lens.

U₀ = object distance from objective lens

d = minimum distance of distinct vision (25 cm)

f_e = Focal length of eyepiece

Now putting the values in equation (4)

m = )

m = 88.

Hence the magnification power of the microscope is 88 and it is a dimensionless quantity as obvious from calculation.

Given:

Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as,

…(1)

Where,

f₀ = focal length of objective lens

f_e = focal length of eyepiece

By putting the values in equation (1) we get,

⇒ m = 144/6

⇒ m = 24

The separation between the objective lens and the eyepiece is calculated as:

d = fo + fe

⇒ d = 144 + 6

⇒ d = 150cm

Hence 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm.

NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.

Given:

Focal length of the objective lens, fo = 15m = 1500 cm
Focal length of the eyepiece, fe = 1.0 cm

(a) The angular magnification of a telescope is given as:

α = fo × fe

⇒ α = 1500 × 1

⇒ α = 1500
Hence, the angular magnification of the given refracting telescope is 1500.

(b) Diameter of the moon, d = 3.48 × 10⁶ m
Radius of the lunar orbit, r₀ = 3.8 × 10⁸ m
Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

The relation between the given variables is,

…(2)

Where,

d = radius of the moon

r₀ = radius of lunar orbit or the distance of moon from earth.

d^' =

d′ = 13.74 cm

Hence the diameter of the image of moon formed by telescope is 13.74 cm.

Given:

(a) For a concave mirror, the focal length (f) is negative.
Therefore f < 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we can write the lens formula as:

…(1)

The object lies between f and 2f.

2f < u < f

⇒ 1/2f < 1/u < 1/f

⇒ −1/2f < −1/u < −1/f

0 …(ii)

Using equation (1), we get:

1/2f < 1/v < 0

1/v is negative, i.e., v is negative.

1/2f < 1v

2f > v

-v > -2f
Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we have the mirror formula:

…(3)

Using equation (2), we can conclude that:

1/v < 0

v > 0

Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, irrespective of the object distance.

(c) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative,
Therefore u < 0
For image distance v, we have the mirror formula:

⇒

But we have u < 0

Therefore 1/v > 1/f

v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.

Therefore f < 0

When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
It is placed between the focus (f) and the pole.

Therefore f > u > 0

1/f < 1/u < 0

1/f − 1/u < 0
For image distance v, we have the mirror formula:

Therefore 1/v < 0

v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:

1/u > 1/v

v > u
Magnification, m = v/u > 1
Hence, the formed image is enlarged

Given:

Actual depth of the pin, d = 15 cm
Let apparent depth of the pin = d’ cm
Refractive index of glass, μ = 1.5

Ratio of actual depth to the apparent depth is equal to the refractive index of glass,
i.e.

μ = d / d′

⇒ d’ = d × μ

⇒ d^’ = 15 × 1.5 = 10cm

The distance at which the pin appears to be raised D,

D = d’-d

D = 15-10 = 5cm

So the pin appears to be raised by 5 cm.
For a small angle of incidence, this distance does not depend upon the location of the slab.

Given:

(a) Refractive index of the glass fibre,

Refractive index of the outer covering of the pipe, μ₂ = 1.44
Angle of incidence = i°
Angle of refraction = r°
Angle of incidence at the interface = i^’
The refractive index (μ) of the core interface is given as:

μ = …(1)

Where,

�₂ = refractive index of outer covering

�₁ = refractive index of inner covering

i^’ = angle of incidence at interface

From equation (1)

sin i^’ = �₁/ �₂

⇒ sin i^’ = 1.44/1.68

⇒ i^' = 59°

For the critical angle, total internal reflection takes place only when i > i’,

i.e., i > 59°
Maximum angle of reflection,

r_max = 90°−i′

⇒ r_max = 90°−59° = 31°
Let, imax be the maximum angle of incidence.
The refractive index at the air − glass interface, μ₁ = 1.68

⇒

⇒ sin i_max = �₁ × sin r_max

sin i_max = 1.68 × 0.5150 = 0.8652

Therefore i_max = sin^-1(0.86)

i_max = 60°

Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ1 = Refractive index of air = 1
For the angle of incidence i = 90°, we can write Snell’s law at the air − pipe interface as:
sin i /sin r = μ

Where,

i = angle of incidence

r = angle of refraction

sin i/ sin r = 1.68

sinr = sin90°/1.68

Sin r = 1/1.68

r = 36.5°

i^’ = 90° - r

⇒ i′ = 90°−36.5°

⇒ i^’ = 53.5°

Since i’ > r, all incident rays will suffer total internal reflection.

Yes,

Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

No,

When light rays diverge, a virtual image is formed. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result the fisherman will appear to be taller.

Yes, decrease, the apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.

Yes,

The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces, so that the diamond sparkles.

Given:

Distance between the object and the image, d = 3 m

Maximum focal length of the convex lens = f_max
For real images, the maximum focal length is given by,

f_max = d/4 …(1)

Where,

f_max = maximum focal length

d = distance between object and image

By putting the values in equation (1) , we get,

f_max = 0.75 m

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Given:

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:

f = …(1)

Where,

f = focal length

D = distance between image and the object.

d = Distance between two locations of the convex lens.

By putting the values in equation (1) we get,

⇒ f = 21.39cm
Therefore, the focal length of the convex lens is 21.39 cm.

Given:

Focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = −20 cm
Distance between the two lenses, d = 8.0 cm
When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:

…(1)

Where,
u₁ = Object distance
v₁ = Image distance

f₁ = Focal length of convex lens.

By putting the values in equation (1), we get,

By solving the above equation we get,

v₁ = 30cm
This image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:

…(2)

Where,
u₂ = Object distance

u₂ = image distance- distance between lenses
⇒ u₂ = (30 − d) = 30 − 8 = 22 cm

v₂ = image distance

f₂ = focal length of concave lens

Now, putting the values in equation (2) we get,

⇒

Therefore v₂ = −220cm

The parallel incident beam appears to diverge from a point that
is (220−d₂) = (220−4) = 216cm from the centre of the combination of the two lenses.
When the parallel beam of light is incident, from the left, on the concave lens I)
According to the lens formula, we have:

…(3)

From equation (3) we get,

Where,
u₂ = Object distance = −∞

v₂ = Image distance

by putting the values in above equation we get,

Therefore v₂ = −20cm

The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:

…(4)

Where,
u₁ = Object distance
⇒ u₁ = −(20 + d) = −(20 + 8) = −28 cm
v₁ = Image distance

Therefore v₂ = −420cm

Hence, the parallel incident beam appear to diverge from a point that is (420 − 4) = 416 cm from the left of the centre of the combination of the two lenses. The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

Given:

Focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = −20 cm
Distance between the two lenses, d = 8.0 cm
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|u1| = 40cm

According to the lens formula:

…(1)

Where,
u₁ = Object distance
v₁ = Image distance

f₁ = Focal length of convex lens.

By putting the values in equation (1), we get,

By solving the above equation we get,

v₁ = 120 cm
Magnification, m = |v₁|/|u₁| = 120/40 = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.

According to the lens formula:

…(2)

Where,
u₂ = Object distance

U₂ = + (120 − 8) = 112 cm.
v₂ = Image distance

f₂ = Focal length of convex lens.

By putting the values in equation (2), we get,

By solving the above equation, we get,

v_{2 =} -2240/92

magnification, m’ = |v₂/u₂| = 2240/92 × 1/112 = 20/92

Hence, the magnification due to the concave lens is 2092.
The magnification produced by the combination of the two lenses is calculated as:

M = m × m^’ = 0.652
The magnification of the combination is given as:

h₂/h₁ = 0.652

h₂ = h₁ × 0.652
Where,
h₁ = Object size = 1.5 cm
h₂ = Size of the image
Therefore h₂ = 0.652 × 1.5 = 0.98cm

Hence, the height of the image is 0.98 cm.

Given:

Angle of prism, A = 60°
Refractive index of the prism, μ = 1.524

Emergent angle, e = 90°

i₁ = Incident angle
r₁ = Refracted angle
r₂ = Angle of incidence at the face AC
According to Snell’s law, for face AC, we can have:

…(1)

From equation(1) we have,

⇒ sin r₂ =

Putting the values in the above equation we find that,

r₂ = sin

r₂ = 41°

We know that, refracted angle r₁ is given by,

r₁ = A- r₂

r₁ = 60-41 = 19°

Therefore r₁ = A−r₂ = 60−41 = 19°
According to Snell’s law, we have the relation:

μ =

Sin i₁ = 1.524 × sin19° = 0.496

⇒ i₁ = 29.75°

Therefore, the angle of incidence is 29.75°

(a) Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.
(b) Take the system of the two prisms as suggested in answer (a). Increase the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.

Given:

Least distance of distinct vision, d = 25 cm
Far point of a normal eye, d’ = ∞
Converging power of the cornea,
Least converging power of the eye-lens, Pc = 40D
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60D

Power of the eye-lens is given as:

P = 1/ focal length

f = 1/P × 1/60D

100/60 = 5/3cm
To focus an object at the near point, object distance (u) = −d = −25 cm
Focal length of the eye-lens = Distance between the cornea and the retina = Image
distance
Hence, image distance, v = 53
Applying the lens formula we have:

…(1)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

1/f^’ = 3/5 + 1/25

1/f^’ = (15 + 1)/25 = 16/25cm^-1

Power P’ = 1/f′ × 100

P^’ = 16/25 × 100 = 64D

Therefore Power of the eye-lens = 64 − 40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20D to 24D

Given:

When the eye-lens loses its ability of accommodation, the defect is called presbyopia. In some cases, the eye-lens is normally accommodated by a person undergoing through myopia or hypermetropia. When the eye-balls get elongated from front to back then it is myopia. Hypermetropia, on the other hand occurs when the eye-balls get shortened.

The myopic person uses a spectacle of power, P = -1.0 D

Focal length of the given spectacles, f = 1/P

f = 1/(-1 × 10^-2 )

f = -100 cm

Hence, 100 cm is the far point of the person. He might have a normal near point of 25 cm. The objects placed at infinity produce virtual images at 100 cm, when spectacles are being used. He is able to see the objects placed between 100 cm and 25 cm when he uses the ability of accommodation of the eye-lens.

During old age, the person uses reading glasses of power, P’ = + 2 D

Old age decreases the ability of accommodation. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

The person is not able to see horizontal lines clearly, whereas the vertical lines are distinctly visible. This problem arises due to problem in refracting mechanism of eye specially cornea and eye lens. This condition is called as astigmatism. The curvature of eye lens in the horizontal plane may be abnormal. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Given:

(a) Focal length of the magnifying glass, f = 5 cm

Distance vision has the least distance, d = 25 cm

Closest object distance = u

Image distance, v = -d = -25 cm

Applying the lens formula we have:

…(1)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

⇒ u = -25/6

⇒ u = 4.16 cm

Hence, the closest distance at which the person can read the book is 4.167 cm.

For the object at the longest distance (u’), the image distance (v’) = ∞

Again using equation (1), we get,

By solving the above equation we get,

u^’ = – 1/5

Therefore, u’ = -5 cm

Hence, the farthest distance at which the person can read the book is 5 cm.

(b) Maximum angular magnification is given by the relation:

α_max = d/|u′|

= 25/5

= 5

Hence the maximum magnification is 5.

Given:

Area of each square, A = 1 mm²

Object distance, u = −9 cm

Focal length of a converging lens, f = 10 cm

(a) Applying the lens formula we have:

…(1)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

The lens formula for the image distance v, can be written as:

…(2)

By putting available values in equation (2) we get,

⇒ V₀ = -90 cm

Magnification, m = v/u

m = −90/−9

m = 10

Therefore, area of each square in the virtual image = (10)² A

= 10² x 1 = 100 mm²

= 1 cm²

(b) The lens has a magnifying power of = d/ |u|

= 25/9

= 2.8

(c) The magnification in (a) is not the same as the magnifying power in (b). Both the quantities will be same once the image is formed at the near point (25 cm)

Given:

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

When the image is formed at the near point or at the minimum distance of distinct vision, we get the maximum possible magnification.

Applying the lens formula we have:

…(1)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Putting values in the equation (1) we get,

⇒

⇒ u = -50/7

⇒ u = -7.14 cm

Hence, the lens should be kept 7.14 cm away in order to view the squares distinctly.

Magnification = |v/u|

= 25/7.14 = 3.5

Magnifying power = d/u

= 25/7.14

= 3.5

Therefore, the magnifying power is equal to the magnitude of magnification since the image is formed at the near point (25 cm).

Given:

Area of the virtual image of each square, A = 6.25 mm²

Area of each square, A₀ = 1 mm²

Hence, the linear magnification of the object can be calculated as:

…(1)

Where, m = magnification

A = Area of virtual image

A₀ = Area of each square

Putting values in equation (1) we get,

m =

⇒ m = 2.5

Also,

m = Image distance(v)/Object distance(u)

m = v/u

Therefore, v = m × u

= 2.5u …(2)

The magnifying glass has a focal length of, f = 10 cm

Applying the lens formula for lens we have:

…(3)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Putting values in the equation (3) we get,

⇒

⇒ u = -(1.5 × 10) / 2.5

Therefore, u = -6 cm

And v = 2.5u

⇒ v = 2.5 × 6

⇒ v = -15 cm

The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.

The angular size of the image is equal to the angular size of the object, but still the angular size of image is equal to angular size of the object. The objects placed closer than the least distance of distinct vision (i.e., 25 cm) can be viewed through a magnifying glass. A closer object causes a larger angular size. A magnifying glass provides angular magnification. The object cannot be placed closer to the eye, without magnification. With magnification, the object can be placed much closer to the eye.

Yes, the angular magnification changes. The angular magnification decreases a little, when the distance between the eye and a magnifying glass is increased. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

Making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length. Thus, the focal length of a convex lens cannot be decreased by a greater amount.

A compound microscope produces an angular magnification of [(25/f_e) + 1]

where,

f_e = focal length of the eyepiece

It can be inferred that if f_e is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens of a compound microscope is given as

1/(|u₀|f₀)

Where, u_o = object distance for the objective lens

f_o = focal length of the objective

The magnification is large when u_o > f_o. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since, u_o is small, f_o will be even smaller. Therefore, f_e and f_o are both small in the given condition.

We are unable to collect much refracted light when we place our eyes too close to the eyepiece of a compound microscope. As a result, there is substantial decrement in the field of view. Hence, the clarity of the image gets blurred. The eye-ring attached to the eyepiece gives the best position for viewing through a compound microscope. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

Given:

Focal length of the objective lens, = 1.25 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

…(1)

By putting the values in equation 1, we get

⇒ m = (1 + 25/5)

⇒ m = 6

The angular magnification of the objective lens (m₀) is related to m_e as:

m_o × m_e = m

⇒ m_o = m/ me

⇒ m₀ = 30/6

⇒ m₀ = 5

We also have the relation:

From the available data we find that,

5 = v₀/(-u₀)

Therefore, v_o = -5u_o ………. (2)

Applying the lens formula for the objective lens:

…(3)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Plugging the values in equation (3)

⇒ u₀ = (-6/5) × 1.25

⇒ u₀ = -1.5 cm

By putting the value of u₀ in equation (2) we can find v₀,

v₀ = -5 × -1.5

v₀ = 7.5 cm

1fo = 1vo–1uo 11.25 = 1−5uo−1uo

= −65uo

Therefore, u_o = −65 × 1.25

= -1.5 cm

And v_o = -5u_o

= -5 x (-1.5)

= 7.5 cm

The object should be placed 1.5 cm away from the objective lens to obtain the required magnification.

Applying the lens formula for the eyepiece:

Applying the lens formula for the objective lens:

…(3)

Where, f_e = focal length of the objective lens

v_e = Image distance for eyepiece = -25cm

u₀ = Distance of object from objective

Plugging the values in equation (3)

⇒ u_e = (-6/25)

⇒ u_e = -4.17 cm

Separation between the objective lens and the eyepiece d = |u_e| + |v_o|

d = 4.17 + 7.5

d = 11.67 cm

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.

Given:

Focal length of the objective lens, = 140 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

(a) When the telescope is in normal adjustment, its magnifying power is given as:

m = f_o/f_e …(1)

By putting the given values in equation (1), we get,

m = 140/5

⇒ m = 28

(b) When the final image is formed at d, the magnifying power m, of the telescope is given as:

…(2)

By putting the values in equation (2), we get,

⇒ m = 28 × (1 + 0.2)

⇒ m = 33.6

Hence, the magnification in case (a) is 28 and in case (b) is 33.6.

Given:

Focal length of the objective lens, f₀ = 140 cm

Focal length of the eyepiece, f_e = 5 cm

In normal adjustment, the separation between the objective lens and the eyepiece

(The rays striking the objective lens are coming from infinity hence they are converged at the focus by objective lens, a small image is formed at focus of objective, the eyepiece magnifies the image at it focus, hence the length of telescope or distance between the objective and eyepiece is f_e + f_o)

d = f_o + f_e

⇒ d = 140 + 5

⇒ d = 145 cm

Given:

Focal length of the objective lens, f₀ = 140 cm

Focal length of the eyepiece, f_e = 5 cm

Height of the tower, h₁ = 100 m

Distance of the tower (object) from the telescope, u = 3Km = 3000 m

The angle subtended by the tower at the telescope is given as:

θ = h₁/ u

⇒ θ = 100/3000

⇒ θ = 1/30 rad

The angle subtended by the image produced by the objective lens is given as:

θ = h₂/ fo

⇒ θ = h₂ /140 rad

Where, h₂ = height of the image of the tower formed by the objective lens

1/30 = h2 /140

Therefore,

h₂ = 140/30

⇒ h₂ = 4.7 cm

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

Given:

Focal length of the objective lens, f₀ = 140 cm

Focal length of the eyepiece, f_e = 5 cm

Image is formed at a distance, d = 25cm

The magnification of the eyepiece is given by the relation:

m = 1 + d/f_e

⇒ m = 1 + 25/5

⇒ m = 6

Height of the final image H, is given by,

H = m × h₂ = 6 × 4.7 = 28.2 cm

Hence, the height of the final image of the tower is 28.2 cm.

Given:

Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of the objective mirror, R₁ = 220 mm

Therefore, focal length of the objective mirror, f₁ = 0.5 × R1 = 110 mm

Radius of curvature of the secondary mirror, R₂ = 140 mm

Therefore, focal length of the secondary mirror, f₂ = 0.5 × R2 = 70 mm

The image of an object placed at infinity, formed by the objective mirror, acts as a virtual object for the secondary mirror.

Let the virtual object distance for the secondary mirror be u,

u = f₁ – d …(1)

u = 110 – 20

u = 90 mm

By applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:

…(2)

By putting the values in equation (2), we get,

v = 315 mm

Hence, the final image will be formed 315 mm away from the secondary mirror.

NOTE: Cassegrainian telescopes are one of the most widely used telescope of current times. It is a combination of a parabolic primary mirror and secondary hyperbolic mirror.

Given:

Angle of deflection, θ = 3.5°

Distance of the screen from the mirror, D = 1.5 m

The reflected rays get deflected by an amount twice the angle of deflection i.e.,

2θ = 7.0°

The displacement (d) of the reflected spot of light on the screen is given as:

tan2θ = d/1.5

(Using trigonometric ratio)

d = 1.5 x tan7^o

d= 0.184 m = 18.4 cm

Hence, the displacement of the reflected spot of light is 18.4 cm

Given:

Focal length of the convex lens, f₁ = 30 cm

The liquid acts as a mirror. Focal length of the liquid = f₂

Focal length of the system (convex lens + liquid), f = 45 cm

Here we have combination of lenses,

For a pair of optical systems placed in contact, the equivalent focal length is given as:

…(1)

By putting the values in equation (1), we can evaluate f_2,

f₂ = – 90cm

Hence, the focal length of liquid is -90 cm

Let the refractive index of the lens be and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.

R can be obtained using the relation:

= (μ₁−1)() …(2)

Putting the values in equation (2), we get,

130 = (1.5 – 1) (2/R)

By the solving the above equation we get,

R = 30cm

Let μ₂ be the refractive index of the liquid.

Radius of curvature of the liquid on the side of the plane mirror = ∞

Radius of curvature of the liquid on the side of the lens, R = −30 cm

The value of μ₂ can be calculated using the relation:

= (μ₂−1) × () …(3)

By plugging the values in equation 3, we get,

μ₂ – 1 = 1/3

Therefore, μ₂ = 1.33

Hence, the refractive index of the liquid is 1.33.