back to directory
NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

A convex lens A of focal length 20 cm20\text{ cm} and a concave lens B of focal length 5 cm5\text{ cm} are kept along the same axis with the distance dd between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance dd in cm will be:

A

25

B

15

C

30

D

50

Step-by-Step Solution

  1. Analysis of Lens A (Convex): A parallel beam of light incident on a convex lens converges at its principal focus. Therefore, lens A attempts to form an image at a distance vA=fA=+20 cmv_A = f_A = +20\text{ cm} from its optical center.
  2. Analysis of Lens B (Concave): For the light rays emerging from lens B to be parallel, the object for lens B must be located at its focal point (or effectively, the rays entering B must be directed towards its focus on the other side).
  3. Afocal System Condition: When a combination of lenses receives parallel light and emits parallel light, the separation distance dd is the algebraic sum of their focal lengths: d=fA+fBd = f_A + f_B.
  4. Calculation: fA=+20 cmf_A = +20\text{ cm} (Convex) fB=5 cmf_B = -5\text{ cm} (Concave)
  • d=20+(5)=15 cmd = 20 + (-5) = 15\text{ cm}.
  1. Alternative Method (Lens Formula): The image formed by A acts as a virtual object for B. Let the distance be dd. Object distance for B: uB=+(20d)u_B = +(20 - d) (since the rays are converging towards a point 20 cm20\text{ cm} from A). Final image distance for B: vB=v_B = \infty (parallel beam).
  • Using 1v1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} for lens B: 1120d=15\frac{1}{\infty} - \frac{1}{20 - d} = \frac{1}{-5} 0120d=150 - \frac{1}{20 - d} = -\frac{1}{5} 120d=15\frac{1}{20 - d} = \frac{1}{5} 20d=5    d=15 cm20 - d = 5 \implies d = 15\text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSconvexlengthconcavelengthdistance

More RAY OPTICS AND OPTICAL INSTRUMENTS Questions

View all

A horizontal ray of light is incident on a right-angled prism with a prism angle of $6^{\circ}$. If the refractive index of the material of the prism is 1.5, then the angle of emergence will be:

A.$9^{\circ}$
B.$10^{\circ}$
C.$4^{\circ}$
D.$6^{\circ}$
MediumSolve

A concave lens with a focal length of $-25 \text{ cm}$ is sandwiched between two convex lenses, each with a focal length of $40 \text{ cm}$. The power (in diopters) of the combined lens system would be:

A.5
B.9
C.1
D.0.01
MediumSolve

The refracting angle of a prism is $A$, and the refractive index of the material of the prism is $\cot(A/2)$. The angle of minimum deviation is:

A.$180^\circ - 3A$
B.$180^\circ - 2A$
C.$90^\circ - A$
D.$180^\circ + 2A$
MediumSolve

A small telescope has an objective of focal length $140\text{ cm}$ and an eyepiece of focal length $5.0\text{ cm}$. The magnifying power of the telescope for viewing a distant object is:

A.28
B.17
C.32
D.34
EasySolve

An astronomical telescope has an objective and eyepiece of focal lengths $40\text{ cm}$ and $4\text{ cm}$ respectively. To view an object $200\text{ cm}$ away from the objective, the lenses must be separated by a distance of:

A.46.0 cm
B.50.0 cm
C.54.0 cm
D.37.3 cm
MediumSolve

A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices $\mu_1$ and $\mu_2$ and $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

A.$\frac{R}{2(\mu_1+\mu_2)}$
B.$\frac{R}{2(\mu_1-\mu_2)}$
C.$\frac{R}{\mu_1-\mu_2}$
D.$\frac{2R}{\mu_2-\mu_1}$
MediumSolve

A double convex lens has a focal length of 25 cm. The radius of curvature of one of the surfaces is double of the other. What would be the radii if the refractive index of the material of the lens is 1.5?

A.100 cm, 50 cm
B.25 cm, 50 cm
C.18.75 cm, 37.5 cm
D.50 cm, 100 cm
MediumSolve

The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. The refractive index of the material of the lens is:

A.4/3
B.9/8
C.5/3
D.3/2
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →