Questions & explanations
1. According to the Cartesian sign convention for spherical mirrors, which of the following statements is correct?
- A. Distances measured in the direction of the incident light are taken as negative.
- B. Distances measured below the principal axis are taken as positive.
- C. The focal length of a concave mirror is taken as negative.
- D. The object distance (u) is always positive for real objects.
Answer: C. The focal length of a concave mirror is taken as negative.
In the Cartesian sign convention, distances measured in the direction of incident light are taken as positive (so A is false). For heights, those above the principal axis are positive and those below are negative (B false). For a real object placed in front of a mirror, the object distance is negative (D false). The focal length of a concave mirror is negative because its focus is in front of the mirror, opposite to the direction of incident light.
2. A spherical mirror has its reflecting surface curved inwards. The centre of the sphere of which the mirror is a part is called the
- A. Focus
- B. Pole
- C. Centre of curvature
- D. Aperture
Answer: C. Centre of curvature
The centre of curvature of a spherical mirror is the centre of the sphere from which the mirror is cut. For a concave mirror, the reflecting surface is curved inward, and the centre of curvature lies in front of the mirror. The focus (A) is the point where parallel rays converge, the pole (B) is the centre of the reflecting surface, and aperture (D) is the diameter of the mirror.
3. A convex spherical surface of radius of curvature 20 cm separates air (n = 1) and glass (n = 1.5). An object is placed in air at a distance of 30 cm from the surface. Where is the image formed?
- 180 cm on the same side as the object
- 180 cm on the opposite side of the surface
- 90 cm on the same side as the object
- 90 cm on the opposite side of the surface
Answer: 180 cm on the same side as the object
Using the formula for refraction at a single spherical surface: n2/v - n1/u = (n2 - n1)/R. Taking u = -30 cm (object distance, real), R = +20 cm (convex surface facing incident light), n1 = 1, n2 = 1.5, we get 1.5/v - 1/(-30) = (1.5-1)/20 → 1.5/v + 1/30 = 0.025 → 1.5/v = -0.00833 → v = -180 cm. The negative sign indicates a virtual image formed on the same side as the object.
4. In normal adjustment (final image at infinity), the magnifying power of a compound microscope is expressed as (where L is the tube length, f_o and f_e are the focal lengths of objective and eyepiece respectively, and D is the least distance of distinct vision):
- M = (L/f_o) × (D/f_e)
- M = (L/f_o) × (1 + D/f_e)
- M = (1 + L/f_o) × (D/f_e)
- M = (L/f_e) × (D/f_o)
Answer: M = (L/f_o) × (D/f_e)
In normal adjustment, the final image is at infinity. The magnification contributed by the objective is L/f_o (when the objective forms the image at its second focal plane), and the eyepiece provides an angular magnification of D/f_e. Hence the total magnifying power is (L/f_o) × (D/f_e). The other options correspond to different adjustments or incorrect combinations.
5. A concave lens of focal length 20 cm forms an image of an object placed 30 cm from the lens. The image is:
- virtual, erect, and diminished
- real, inverted, and same size
- virtual, inverted, and enlarged
- real, erect, and diminished
Answer: virtual, erect, and diminished
Using the thin lens formula with f = -20 cm and u = -30 cm: 1/f = 1/v - 1/u → 1/v = 1/f + 1/u = -1/20 - 1/30 = -5/60 → v = -12 cm. The negative v indicates a virtual image. Magnification m = v/u = (-12)/(-30) = +0.4, positive (erect) and less than 1 (diminished). Concave lenses always produce virtual, erect, and diminished images.
6. Which of the following correctly represents Snell's law of refraction?
- n1 sin i = n2 sin r
- n1 sin r = n2 sin i
- n1 / n2 = sin i / sin r
- n1 n2 = sin i sin r
Answer: n1 sin i = n2 sin r
Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant and equal to the refractive index of the second medium with respect to the first. This is expressed as n1 sin i = n2 sin r, where n1 and n2 are the refractive indices of the two media.
7. Which of the following statements correctly distinguishes the focal length requirements of objective and eyepiece in a compound microscope and an astronomical telescope?
- Microscope: objective short, eyepiece short; Telescope: objective long, eyepiece short
- Microscope: objective short, eyepiece long; Telescope: objective long, eyepiece short
- Microscope: objective long, eyepiece short; Telescope: objective short, eyepiece long
- Microscope: objective long, eyepiece long; Telescope: objective short, eyepiece short
Answer: Microscope: objective short, eyepiece short; Telescope: objective long, eyepiece short
For high magnification in a compound microscope, both objective and eyepiece have short focal lengths. In an astronomical telescope, the objective has a long focal length (to reduce the angular size of the image) while the eyepiece has a short focal length (to provide high angular magnification).
8. A convex mirror has a focal length of 10 cm. An object is placed 30 cm in front of it. What is the image distance and its nature?
- A. +7.5 cm, virtual and erect
- B. -7.5 cm, real and inverted
- C. +15 cm, virtual and erect
- D. -15 cm, real and inverted
Answer: A. +7.5 cm, virtual and erect
For a convex mirror, f = +10 cm, u = -30 cm. Mirror formula: 1/v = 1/f - 1/u = 1/10 - 1/(-30) = 1/10 + 1/30 = 4/30 = 2/15, so v = +7.5 cm. Positive v indicates image behind the mirror, which is virtual and erect. Convex mirrors always form virtual, erect, and diminished images.
9. The lens maker's formula for a thin lens in air is 1/f = (n-1)(1/R1 - 1/R2). What is the focal length of a biconvex lens made of glass (n = 1.5) with radii of curvature 20 cm and 30 cm?
- 24 cm
- 12 cm
- 60 cm
- 30 cm
Answer: 24 cm
For a biconvex lens, the first surface is convex (R1 = +20 cm) and the second surface is concave (R2 = -30 cm). Substituting into the lens maker's formula: 1/f = (1.5-1)(1/20 - 1/(-30)) = 0.5 × (0.05 + 0.03333) = 0.5 × 0.08333 = 0.04167, so f = 1/0.04167 ≈ 24 cm.
10. Which of the following statements about images formed by mirrors is correct?
- A. A convex mirror always forms a real and inverted image.
- B. A concave mirror can form both real and virtual images.
- C. A concave mirror always forms a diminished image.
- D. A convex mirror can form a magnified image.
Answer: B. A concave mirror can form both real and virtual images.
Concave mirrors form real, inverted images when the object is beyond focus, and virtual, erect, magnified images when the object is between focus and pole. Convex mirrors always form virtual, erect, and diminished images. Therefore, only B is correct.
11. Which of the following phenomena is NOT explained by total internal reflection?
- Sparkling of diamond
- Formation of a mirage
- Blue colour of the sky
- Light propagation through optical fibres
Answer: Blue colour of the sky
The blue colour of the sky is due to Rayleigh scattering of sunlight by air molecules, not total internal reflection. The other three options – sparkling of diamond, mirage, and optical fibres – are all applications of total internal reflection.
12. A convex lens of focal length 15 cm forms a real image of an object placed 30 cm from it. What is the magnification produced?
- -1
- +1
- -2
- +0.5
Answer: -1
Using the thin lens formula 1/f = 1/v - 1/u with u = -30 cm and f = +15 cm: 1/15 = 1/v - 1/(-30) → 1/15 = 1/v + 1/30 → 1/v = 1/15 - 1/30 = 1/30 → v = 30 cm. Magnification m = v/u = 30/(-30) = -1. The negative sign indicates an inverted image.