Q. Answer the following questions:

• (i) What is relative refractive index?
• (ii) Define one dioptre of power of a lens.
• (iii) How lateral shift depends on the angle of incidence?
• Ans. (i) Refractive index of a medium with respect to another medium is called relative refractive index.
• (ii) One dioptre is defined as the power of a lens of focal length 1 metre.
• ( iii) Lateral shift is directly proportional to angle of incidence.

Q. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

F ocal length, f = R/2

f = 32/2 cm

f = 16 cm

Q. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Ans. Focal length of convex mirror, f = +15 cm
Object distance, u = – 10 cm

Accdording to themirror of formula, $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ Substituting the values, we get $$\frac{1}{v} + \frac{1}{-10} = \frac{1}{15}$$ $$\frac{1}{v} = \frac{1}{15} + \frac{10}{10}$$ $$\frac{1}{v} = \frac{2+3}{30}$$ $$\frac{1}{v} = \frac{5}{30}$$ $$\frac{1}{v} = \frac{1}{6}$$ $$v = + \space 6 \space cm$$

Thus, the image is for med at a distance of 6 cm from the convex mirror. Since the image is formed behind the convex mirror, so image is virtual and erect.

Q. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m/s .
• Ans. Speed of light in vacuum, c = 3 × 108 m/s
• Refractive index of glass, ng = 1.50

We  have, $$n_g = \frac{c}{v}$$ or $$v = \frac{c}{n_g}$$ $$\therefore \space \space \space \space \space v = \frac{3 × 10^8}{1.5}$$ $$v = 2 × 10^8 m/s$$

Q. Find the power of a concave lens of focal length 2 m.

• Ans. Focal length of concave lens, f = –2 m
• (Since, the focal length of concave lens is negative.)
• P = 1/p=i/f=1/-2
• = – 0.5 D
• Hence, the power of the given concave lens is – 0.5 D.

Q. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

• Ans. Object distance, u = – 25 cm
• Object height, h1 = 5 cm
• Focal length, f = +10 cm

Using the lens formula, $$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$$ Substituting the values, We get $$\frac{1}{v} – \frac{1}{-25} = \frac{1}{10}$$ $$\frac{1}{v} + \frac{1}{25} = \frac{1}{10}$$ $$\frac{1}{v} = \frac{3}{50}$$ $$\frac{1}{v} = \frac{50}{3}$$ $$v = 16.67cm$$ The positive value of v shows that the image is formed at the other side of the lens.
Magnification, $$m = \frac{h_2}{h_1} = \frac{v}{u} = \frac{16.67}{-25} = -0.66$$
The negative sign shows that the image is real and formed behing the lens.
Magnification, $$m = \frac{h_2}{h_1}$$ $$h_2 = m × h_1 = (-0.66) × 5$$ $$= -3.3 \space cm.$$

The negative value of image height indicates that the image formed is inverted. The position, size and nature of image are shown in the following ray diagram.

Q. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

• Ans. Focal length of concave lens, f = – 15 cm
• Image distance, v = – 10 cm
• A ccording to the lens formula,

$$\frac{1}{v}-\frac{1}{u} = \frac{1}{f}$$
Substituting the values, we get $$\frac{1}{-10} – \frac{1}{u} = \frac{1}{-15}$$ $$-\frac{1}{u} = \frac{-1}{15} + \frac{-1}{10}$$ $$-\frac{1}{u} = \frac{-2+3}{30}$$ $$-\frac{1}{u} = \frac{1}{30}$$ $$u = -30 \space cm$$

The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.

Q. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

• Ans. Since the focal length of mirror is 15 cm, the range of object distance = 0 cm to 15 cm. A concave mir ror gives an erect image when an object is placed between its pole (P) and the principal focus (F). So, the image formed will be virtual, erect, and magnified, as shown in the given figure.

Q. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

• Ans. The convex lens will form complete image of an object, even if the one half is covered with black paper, because light rays can still pass through the optical centre of convex lens. We can verify this by obtaining image of any distant object on a screen by half covered convex lens. This can be more clear by the ray diagram given alongside.