JEE Physics Ray Optics and Optical Instruments MCQs Set 06

Physics Objective Questions and Answers: Ray Optics and Optical Instruments

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Multiple Choice Questions

Question 1: A convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together. The resulting power is
(a) +6.5 D
(b) -6.5 D
(c) +7.5 D
(d) -0.75 D
Show Answer & Explanation

Answer: (d) -0.75 D

Explanation:
1. \( P=100/80-100/50=1.25-2=-0.75 \) D.
2. The combination is diverging.

Teacher's Note:
1. Convert focal lengths to dioptres.

Question 2: A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Its focal length will
(a) Become zero
(b) Become infinite
(c) Become small but non-zero
(d) Remain unchanged
Show Answer & Explanation

Answer: (b) Become infinite

Explanation:
1. The power depends on \( (\mu_g/\mu_l-1) \), which becomes zero.
2. So the focal length becomes infinite.

Teacher's Note:
1. The lens vanishes optically.

Question 3: An equiconvex lens is cut into two halves (i) along a plane containing its principal axis, and (ii) along the plane perpendicular to the principal axis through the optical centre. Let f, f', f'' be the focal lengths of the complete lens, of each half in case (i) and of each half in case (ii) respectively. The correct statement is
(a) f' = f, f'' = 2f
(b) f' = 2f, f'' = f
(c) f' = f, f'' = f
(d) f' = 2f, f'' = 2f
Show Answer & Explanation

Answer: (a) f' = f, f'' = 2f

Explanation:
1. In case (i) each half has the same curved surfaces and the same radii, so f' = f.
2. In case (ii) each half is a plano-convex lens with one curved surface, so its focal length doubles, f'' = 2f.

Teacher's Note:
1. Power depends on the curved surfaces.

Question 4: An object and its real image are located at distances 25 cm and 40 cm respectively from the two principal foci of a convex lens. The linear magnification of the image is near
(a) +1.3
(b) -1.3
(c) +1.8
(d) -1.8
Show Answer & Explanation

Answer: (b) -1.3

Explanation:
1. \( f^2=x_1x_2=25\times40=1000 \), so \( f=31.6 \) cm.
2. \( m=-f/x_1=-31.6/25\approx-1.3 \).

Teacher's Note:
1. Real image from a convex lens is inverted.

Question 5: Lenses of dispersive powers 0.024 and 0.036 form an aberration-free (achromatic) combination. Their focal lengths will be
(a) 30 cm, -40 cm
(b) 30 cm, -45 cm
(c) 10 cm, 30 cm
(d) 20 cm, -35 cm
Show Answer & Explanation

Answer: (b) 30 cm, -45 cm

Explanation:
1. Condition \( \omega_1/f_1+\omega_2/f_2=0 \) means the focal lengths have opposite signs and magnitudes in the ratio \( 0.024:0.036=2:3 \).
2. Only 30 cm and -45 cm satisfy this.

Teacher's Note:
1. Achromatic combination has one convex and one concave lens.

Question 6: An achromatic convergent doublet of two lenses in contact has a power of +2 D. The convex lens has a power of +5 D. The ratio of the dispersive powers of the convergent and divergent lenses is
(a) 2 : 5
(b) 3 : 5
(c) 5 : 2
(d) 5 : 3
Show Answer & Explanation

Answer: (b) 3 : 5

Explanation:
1. The concave lens has power \( 2-5=-3 \) D.
2. For achromatism \( \omega_1P_1+\omega_2P_2=0 \), so \( \omega_1/\omega_2=3/5 \).

Teacher's Note:
1. Greater power needs smaller dispersive power.

Question 7: The dispersive powers of the materials of two lenses are in the ratio 4 : 3. The achromatic combination of these lenses in contact is a convex lens of focal length 60 cm. The focal lengths of the component lenses are
(a) -20 cm and 25 cm
(b) 20 cm and -25 cm
(c) -15 cm and 40 cm
(d) 15 cm and -20 cm
Show Answer & Explanation

Answer: (d) 15 cm and -20 cm

Explanation:
1. \( f_1:f_2=-\omega_1:\omega_2=-4:3 \).
2. With \( 1/f_1+1/f_2=1/60 \) the solution is \( f_1=-20 \) cm for the material of larger dispersive power and \( f_2=+15 \) cm.

Teacher's Note:
1. The convex lens has the lower dispersive power.

Question 8: Two lenses of focal lengths +10 cm and -15 cm when in contact behave like a convex lens. They will have zero longitudinal chromatic aberration if their dispersive powers are in the ratio
(a) +3/2
(b) +2/3
(c) -3/2
(d) -2/3
Show Answer & Explanation

Answer: (b) +2/3

Explanation:
1. \( \omega_1/f_1+\omega_2/f_2=0 \) gives \( \omega_1/\omega_2=-f_1/f_2=-10/(-15) \).
2. \( \omega_1/\omega_2=2/3 \).

Teacher's Note:
1. Dispersive powers are positive; the signs of f carry the minus.

Question 9: The refractive indices of violet, yellow and red light for a material of a lens are 1.66, 1.64 and 1.62. If the mean focal length of the lens is 10 cm, the chromatic aberration between violet and red will be
(a) 0.625 cm
(b) 0.125 m
(c) .02 cm
(d) 0 cm
Show Answer & Explanation

Answer: (a) 0.625 cm

Explanation:
1. \( f_r-f_v=f\,\omega=f(\mu_v-\mu_r)/(\mu_y-1) \).
2. \( =10\times0.04/0.64=0.625 \) cm.

Teacher's Note:
1. Axial chromatic aberration is f times omega.

Question 10: The focal length of a convex lens is 2.5 cm. Its magnifying power for the minimum distance of distinct vision (25 cm) is
(a) 25
(b) 52
(c) 11
(d) 1.1
Show Answer & Explanation

Answer: (c) 11

Explanation:
1. \( M=1+D/f=1+25/2.5=11 \).
2. The image is formed at the near point.

Teacher's Note:
1. M = 1 + D/f.

Question 11: An astronomical telescope of magnifying power 8 is made using two lenses spaced 45 cm apart. The focal lengths of the lenses are
(a) F = 40 cm, f = 5 cm
(b) F = 8 cm, f = 5 cm
(c) F = 5 cm, f = 47 cm
(d) F = 20 cm, f = 5 cm
Show Answer & Explanation

Answer: (a) F = 40 cm, f = 5 cm

Explanation:
1. \( F+f=45 \) and \( F/f=8 \), so \( f=5 \) cm and \( F=40 \) cm.
2. These match option 1.

Teacher's Note:
1. Normal adjustment: L = fo + fe.

Question 12: The magnifying power of the objective of a compound microscope is 7. If the magnifying power of the microscope is 35, the magnifying power of the eyepiece will be
(a) 245
(b) 5
(c) 28
(d) 42
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. \( M=M_o\times M_e \), so \( M_e=35/7=5 \).
2. The total is the product.

Teacher's Note:
1. Magnification multiplies.

Question 13: An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 10 cm. When the final image is formed at the least distance of distinct vision, the magnifying power of the telescope is
(a) -10
(b) -11
(c) -14
(d) -15
Show Answer & Explanation

Answer: (c) -14

Explanation:
1. \( M=-\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)=-10(1+0.4) \).
2. \( M=-14 \).

Teacher's Note:
1. Near point setting has a larger magnification.

Question 14: If the tube length of an astronomical telescope is 105 cm and the magnifying power is 20 for normal setting, the focal length of the objective is
(a) 100 cm
(b) 10 cm
(c) 20 cm
(d) 25 cm
Show Answer & Explanation

Answer: (a) 100 cm

Explanation:
1. \( f_o+f_e=105 \) and \( f_o/f_e=20 \), so \( f_e=5 \) cm and \( f_o=100 \) cm.
2. Normal setting means the final image is at infinity.

Teacher's Note:
1. Tube length = sum of focal lengths.

Question 15: The least distance of distinct vision is 25 cm. The magnifying power of a simple microscope of focal length 5 cm, when the final image is formed at the minimum distance of distinct vision, is
(a) 1/5
(b) 5
(c) 1/6
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. \( M=1+D/f=1+25/5=6 \).
2. The image is at the near point.

Teacher's Note:
1. 1 + D/f for the near-point image.

Question 16: In a compound microscope, the intermediate image in normal use is
(a) Virtual, erect and magnified
(b) Real, erect and magnified
(c) Real, inverted and magnified
(d) Virtual, inverted and magnified
Show Answer & Explanation

Answer: (c) Real, inverted and magnified

Explanation:
1. The objective forms a real, inverted and magnified image of the object.
2. The eyepiece then acts as a magnifier on it.

Teacher's Note:
1. The final image is virtual and inverted.

Question 17: The focal lengths of the objective and eyepiece of a telescope are 100 cm and 2 cm. The moon subtends an angle of 0.5\( ^\circ \) at the eye. If it is looked at through the telescope, the angle subtended by the moon's image will be
(a) \( 100^\circ \)
(b) \( 50^\circ \)
(c) \( 25^\circ \)
(d) \( 10^\circ \)
Show Answer & Explanation

Answer: (c) \( 25^\circ \)

Explanation:
1. \( M=f_o/f_e=50 \).
2. Angle \( =50\times0.5^\circ=25^\circ \).

Teacher's Note:
1. Angular magnification of the telescope.

Question 18: Four convex lenses L1, L2, L3, L4 of focal lengths 2, 4, 6 and 8 cm are available. Two of them form a telescope of length 10 cm and magnifying power 4. The objective and eye lenses are
(a) L2, L3
(b) L1, L4
(c) L3, L2
(d) L4, L1
Show Answer & Explanation

Answer: (d) L4, L1

Explanation:
1. \( f_o+f_e=10 \) and \( f_o/f_e=4 \) give \( f_o=8 \) cm and \( f_e=2 \) cm.
2. So the objective is L4 and the eyepiece is L1.

Teacher's Note:
1. Objective has the larger focal length.

Question 19: The magnifying power of a simple microscope is 6. The focal length of its lens in metres will be, if the least distance of distinct vision is 25 cm and the image is formed at the near point,
(a) 0.05
(b) 0.06
(c) 0.25
(d) 0.12
Show Answer & Explanation

Answer: (a) 0.05

Explanation:
1. \( 6=1+25/f \) gives \( f=5 \) cm.
2. \( f=0.05 \) m.

Teacher's Note:
1. Convert cm to m.

Question 20: An astronomical telescope has a magnifying power 10. The focal length of the eyepiece is 20 cm. The focal length of the objective is
(a) 2 cm
(b) 200 cm
(c) \( \tfrac12\ \text{cm} \)
(d) \( \tfrac1{200}\ \text{cm} \)
Show Answer & Explanation

Answer: (b) 200 cm

Explanation:
1. \( M=f_o/f_e \) gives \( f_o=10\times20=200 \) cm.
2. The objective has the longer focal length.

Teacher's Note:
1. Telescope objective: large f.

Question 21: A telescope has an objective lens of focal length 200 cm and an eyepiece of focal length 2 cm. If this telescope is used to see a 50 m tall building at a distance of 2 km, the height of the image of the building formed by the objective lens is
(a) 5 cm
(b) 10 cm
(c) 1 cm
(d) 2 cm
Show Answer & Explanation

Answer: (a) 5 cm

Explanation:
1. Angle subtended \( =50/2000=0.025 \) rad.
2. Image height \( =f_o\times0.025=200\times0.025=5 \) cm.

Teacher's Note:
1. The objective forms the image in its focal plane.

Question 22: The minimum and maximum distances for clear vision of a healthy eye are
(a) 100 cm and 500 cm
(b) Infinite and 25 cm
(c) 25 cm and 100 cm
(d) 25 cm and infinite
Show Answer & Explanation

Answer: (d) 25 cm and infinite

Explanation:
1. The near point of a normal eye is 25 cm and the far point is at infinity.
2. So clear vision lies between 25 cm and infinity.

Teacher's Note:
1. Normal eye: 25 cm to infinity.

Question 23: A person can see clearly only up to a distance of 25 cm. He wants to read a book placed at a distance of 50 cm. The kind of lens he requires for his spectacles and its power are
(a) Concave, -1.0 D
(b) Convex, +1.5 D
(c) Concave, -2.0 D
(d) Convex, +2.0 D
Show Answer & Explanation

Answer: (c) Concave, -2.0 D

Explanation:
1. The lens must form the image at 25 cm of an object at 50 cm: \( 1/f=1/v-1/u=-1/25+1/50=-1/50 \).
2. \( f=-50 \) cm, a concave lens of power -2 D.

Teacher's Note:
1. This is the correction for a short-sighted eye.

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