Ray Optics and Optical Instruments

Question 11:

Q: Derive the expression for the magnification produced by a concave mirror.

Solution: The magnification $m$ produced by a concave mirror is the ratio of the image height $h_i$ to the object height $h_o$ , i.e.,

$$m = \frac{h_i}{h_o}$$

Using the geometry of the mirror, we know that magnification is also given by:

$$m = -\frac{v}{u}$$

where $v$ is the image distance and $u$ is the object distance. Thus, the magnification is negative for a real and inverted image.

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Question 12:

Q: A concave mirror of focal length 20 cm forms an image of an object 30 cm away from the mirror. What is the magnification of the image?

Solution: Given:

• Focal length $f = 20 \, \text{cm}$ ,
• Object distance $u = -30 \, \text{cm}$ .

Use the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Substitute the known values:

$$\frac{1}{20} = \frac{1}{v} + \frac{1}{-30}$$ $$\frac{1}{v} = \frac{1}{20} + \frac{1}{30}$$ $$\frac{1}{v} = \frac{3 + 2}{60}$$ $$v = \frac{60}{5} = 12 \, \text{cm}$$

Now, calculate the magnification:

$$m = -\frac{v}{u} = -\frac{12}{-30} = 0.4$$

Thus, the magnification is $0.4$ , and the image is diminished and real.

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Question 13:

Q: What is the relationship between the focal length, radius of curvature, and the focal length of a spherical mirror?

Solution: The relationship between the focal length $f$ and the radius of curvature $R$ of a spherical mirror is given by:

$$f = \frac{R}{2}$$

This relationship holds true for both concave and convex mirrors. It is derived by considering the mirror's geometry and the reflection laws.

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Question 14:

Q: A light ray is incident on a convex lens at an angle of 30°. Calculate the refracted angle inside the lens if the refractive index of the lens is $1.5$ .

Solution: Using Snell's law, we can calculate the refracted angle:

$$n_1 \sin i = n_2 \sin r$$

where:

• $n_1 = 1$ (refractive index of air),
• $n_2 = 1.5$ (refractive index of the lens),
• $i = 30^\circ$ (incident angle),
• $r$ is the refracted angle.

Substitute the values into Snell's law:

$$\sin r = \frac{1 \cdot \sin 30^\circ}{1.5} = \frac{0.5}{1.5} = \frac{1}{3}$$ $$r = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ$$

Thus, the refracted angle inside the lens is approximately $19.47^\circ$ .

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Question 15:

Q: Explain the formation of an image by a telescope.

Solution: A telescope works by using two lenses: the objective lens and the eyepiece. The objective lens gathers light from a distant object and forms a real, inverted image at its focus. This image serves as the object for the eyepiece lens, which magnifies it and forms a virtual, erect image at a comfortable viewing distance.

The magnification of the telescope is given by:

$$M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}$$

where $f_{\text{objective}}$ is the focal length of the objective lens, and $f_{\text{eyepiece}}$ is the focal length of the eyepiece.

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Question 16:

Q: What is the working principle of a microscope, and how is it different from a telescope?

Solution: A microscope works on the principle of forming a magnified image of a small object. It uses two lenses: the objective lens and the eyepiece. The objective lens forms a real, inverted image close to the focal point, and the eyepiece further magnifies this image to form a virtual, upright image.

In contrast to a telescope, which is used for distant objects, a microscope is used for nearby objects. The magnification in a microscope is given by:

$$M = \frac{D}{f_{\text{objective}}} + 1$$

where $D$ is the least distance of distinct vision and $f_{\text{objective}}$ is the focal length of the objective lens.

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Question 17:

Q: Calculate the focal length of a lens system consisting of a diverging lens with focal length $-10 \, \text{cm}$ and a converging lens with focal length $20 \, \text{cm}$ , placed in contact.

Solution: For two lenses in contact, the total focal length $f_{\text{total}}$ is given by:

$$\frac{1}{f_{\text{total}}} = \frac{1}{f_1} + \frac{1}{f_2}$$

Substitute $f_1 = -10 \, \text{cm}$ and $f_2 = 20 \, \text{cm}$ :

$$\frac{1}{f_{\text{total}}} = \frac{1}{-10} + \frac{1}{20}$$ $$\frac{1}{f_{\text{total}}} = \frac{-2 + 1}{20} = \frac{-1}{20}$$

Thus, the total focal length is:

$$f_{\text{total}} = -20 \, \text{cm}$$

The combined focal length is $-20 \, \text{cm}$ , indicating a diverging lens.

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Question 18:

Q: A light ray is incident on a plane mirror at an angle of $45^\circ$ . What is the angle between the incident ray and the reflected ray?

Solution: The angle of incidence $i$ is equal to the angle of reflection $r$ . Given that the incident angle is $45^\circ$ , the reflected angle is also $45^\circ$ .

Thus, the angle between the incident ray and the reflected ray is:

$$\text{Angle between incident and reflected ray} = i + r = 45^\circ + 45^\circ = 90^\circ$$

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Question 19:

Q: A person is looking through a magnifying glass. If the focal length of the magnifying glass is $5 \, \text{cm}$ , and the object is placed at a distance of $5 \, \text{cm}$ , calculate the magnification produced.

Solution: The magnification $m$ produced by a lens is given by:

$$m = 1 + \frac{D}{f}$$

where $D$ is the least distance of distinct vision (approximately $25 \, \text{cm}$ ), and $f$ is the focal length of the lens.

Given:

• $D = 25 \, \text{cm}$ ,
• $f = 5 \, \text{cm}$ ,

Substitute the values:

$$m = 1 + \frac{25}{5} = 1 + 5 = 6$$

Thus, the magnification is 6.

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Question 20:

Q: A ray of light strikes the surface of a glass slab at an angle of incidence of $30^\circ$ . If the refractive index of glass is $1.5$ , calculate the angle of refraction inside the glass.

Solution: Using Snell's law:

$$n_1 \sin i = n_2 \sin r$$

where:

• $n_1 = 1$ (refractive index of air),
• $n_2 = 1.5$ (refractive index of glass),
• $i = 30^\circ$ ,
• $r$ is the refracted angle inside the glass.

Substitute the values:

$$\sin r = \frac{1 \cdot \sin 30^\circ}{1.5} = \frac{0.5}{1.5} = \frac{1}{3}$$ $$r = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ$$

Thus, the angle of refraction inside the glass is approximately $19.47^\circ$ .

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Question 21:

Q: Derive the lens formula and magnification formula for a convex lens.

Solution: The lens formula is derived from the geometry of the lens and is given by:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

where:

• $f$ is the focal length,
• $v$ is the image distance,
• $u$ is the object distance.

The magnification $m$ produced by a lens is given by:

$$m = \frac{h_i}{h_o} = -\frac{v}{u}$$

where:

• $h_i$ is the image height,
• $h_o$ is the object height.

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Question 22:

Q: A concave mirror has a focal length of $15 \, \text{cm}$ . An object is placed at a distance of $30 \, \text{cm}$ from the mirror. Where will the image be formed?

Solution: Use the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Substitute:

• $f = 15 \, \text{cm}$ ,
• $u = -30 \, \text{cm}$ .

$$\frac{1}{15} = \frac{1}{v} + \frac{1}{-30}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{2 + 1}{30} = \frac{3}{30} = \frac{1}{10}$$ $$v = 10 \, \text{cm}$$

Thus, the image is formed at a distance of 10 cm from the mirror and is real and inverted.

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Question 23:

Q: Derive the expression for the focal length of a lens in combination.

Solution: For two lenses in combination, the total focal length $f_{\text{total}}$ is given by:

$$\frac{1}{f_{\text{total}}} = \frac{1}{f_1} + \frac{1}{f_2}$$

where:

• $f_1$ and $f_2$ are the focal lengths of the individual lenses.

For more than two lenses, this formula is extended:

$$\frac{1}{f_{\text{total}}} = \sum \frac{1}{f_i}$$

where the summation is taken over all the lenses in the system.

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