Question 1

A convex lens has a focal length of 20 cm. An object is placed at a distance of 60 cm from the lens. What is the nature and position of the image formed?

Accepted Answer

Using lens formula 1/f = 1/u + 1/v, with f = 20 cm and u = 60 cm, we get v = 30 cm; since v is positive, image is real and inverted at 30 cm.

Question 2

A concave mirror has a radius of curvature of 40 cm. An object is placed at the center of curvature. What is the magnification of the image?

Accepted Answer

When object is at center of curvature (u = R = 40 cm), the image also forms at center of curvature with magnification m = -v/u = -1 (real, inverted, same size).

Question 3

The power of a lens is –5 diopters. What is its focal length and type?

Accepted Answer

Power P = 1/f in meters; f = 1/(-5) = -0.2 m = -20 cm, and negative focal length indicates a concave (diverging) lens.

Question 4

A convex lens of focal length 10 cm is used as a simple microscope. An object is placed at 5 cm from the lens. What is the nature of the image formed?

Accepted Answer

When object distance (5 cm) is less than focal length (10 cm), a convex lens forms a virtual, erect, and magnified image on the same side as the object.

Question 5

A concave lens has a focal length of 15 cm. An object is placed at 30 cm from the lens. What is the distance of the image from the lens?

Accepted Answer

Using 1/f = 1/u + 1/v with f = -15 cm and u = 30 cm gives 1/(-15) = 1/30 + 1/v, solving yields v = -10 cm (virtual image).

Question 6

A concave mirror has a focal length of 15 cm. An object is placed at a distance of 30 cm from the mirror. What is the magnification produced?

Accepted Answer

Using mirror formula 1/f = 1/u + 1/v, we get v = -30 cm, so magnification m = -v/u = -(-30)/30 = -1.

Question 7

A convex lens has a power of +5 diopters. What is its focal length?

Accepted Answer

Power P = 1/f (in meters), so f = 1/5 = 0.2 m = 20 cm.

Question 8

Light travels from a denser medium (refractive index 1.5) to a rarer medium (refractive index 1.0). The angle of incidence is 30°. What is the angle of refraction?

Accepted Answer

Using Snell's law: 1.5 × sin(30°) = 1.0 × sin(r), so sin(r) = 0.75, r ≈ 48.6°.

Question 9

The critical angle for total internal reflection for a medium is 45°. What is the refractive index of that medium?

Accepted Answer

At critical angle, n × sin(C) = 1, so n = 1/sin(45°) = 1/0.707 ≈ 1.41 or √2.

Question 10

A thin lens produces a real, inverted image that is twice the size of the object. If the object distance is 20 cm, what is the focal length of the lens?

Accepted Answer

Magnification m = -2 = -v/u, so v = 40 cm; using lens formula 1/f = 1/20 + 1/40 = 3/40, f ≈ 13.3 cm.

Question 11

A convex mirror has a focal length of -10 cm. An object is placed 20 cm in front of it. What is the position of the image formed?

Accepted Answer

Using mirror formula 1/(-10) = 1/20 + 1/v, we get 1/v = -1/10 - 1/20 = -3/20, so v = -6.67 cm (virtual image).

Question 12

The least distance of distinct vision for a normal human eye is approximately:

Accepted Answer

The near point of a normal eye is standardized at 25 cm, which is the minimum distance for clear vision without strain.

Question 13

A concave lens of focal length -20 cm is used as a simple magnifier. What is the maximum angular magnification when the final image is at the near point (25 cm)?

Accepted Answer

Concave lenses diverge light and always form virtual, diminished images; they cannot be used as magnifying glasses.

Question 14

A plane mirror is moved 5 cm away from an object. How much does the distance between the object and its image increase?

Accepted Answer

In a plane mirror, image distance equals object distance; moving mirror 5 cm away increases object distance by 5 cm and image distance by 5 cm, so total separation increases by 10 cm.

Question 15

A concave mirror of focal length 15 cm forms a real, inverted image of height 4 cm from an object of height 2 cm. If the object is placed at a distance of 30 cm from the mirror, what is the image distance?

Accepted Answer

Using mirror formula 1/f = 1/u + 1/v with f = 15 cm and u = 30 cm gives 1/15 = 1/30 + 1/v, so v = 30 cm.

Question 16

A convex lens has a power of +5 diopters. An object is placed 40 cm in front of the lens. Calculate the position of the image formed.

Accepted Answer

Power = +5 D gives f = 20 cm; using lens formula 1/f = 1/u + 1/v with u = 40 cm yields v = 40 cm, but recalculation gives v = 26.7 cm on image side.

Question 17

A compound microscope has an objective lens of focal length 0.5 cm and an eyepiece of focal length 5 cm. If the tube length is 15 cm, what is the magnification produced?

Accepted Answer

Magnification M = (L/f_o) × (D/f_e) where L = 15 cm, f_o = 0.5 cm, f_e = 5 cm, D = 25 cm gives M = (15/0.5) × (25/5) = 30 × 5 = 150×.

Question 18

A plano-convex lens has a radius of curvature of 20 cm for its curved surface. The lens is immersed in water (refractive index 1.33) with its plane surface facing the object. What is the focal length in water? (n_glass = 1.5)

Accepted Answer

Using lens maker's formula in water: 1/f = (n_rel - 1)/R where n_rel = 1.5/1.33 ≈ 1.128, R = 20 cm gives f ≈ 176 cm.

CAPF AC Ray Optics & Optical Instruments

Ray Optics & Optical Instruments — Practice Questions