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Concepts of Physics

Geometrical optics

A converging lens of focal length 15 cm and a converging mirro

Concepts of Physics

Class 11

HC Verma Part I

Chapter 1: Introduction to Physics

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Chapter 2: Physics and Mathematics

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Chapter 3: Rest and Motion: Kinematics

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Chapter 4: The Forces

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Chapter 5: Newton's Law of Motion

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Chapter 9: Centre of Mass, Linear momentum Collision

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Chapter 14: Some Mechanical Properties of Matter

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Chapter 17: Light Waves

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Chapter 18: Geometrical optics

123 questions

Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate (μ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

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Chapter 19: Optical Instruments

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Chapter 21: Speed of Light

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Chapter 22: Photometry

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Question

Hard

Solving time: 10 mins

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

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Updated on: Sep 25, 2022

Book:

Concepts of Physics (HC Verma Part I)

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Text solutionVerified

Given,
Convex lens of focal length

f_{l}

= 15 cm
Concave mirror of focal length

f_{m}

= 10 cm
Distance between mirror and lens = 50 cm
Length of the pin (object length)

h_{0}

= 2.0 cm

As per the question

The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.

Using lens formula,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f _{l}}

\Rightarrow v = \frac{u f _{l}}{u + f _{l}}

Since,

u = - 30 cm

and

f_{1} = 15 cm

S o, v = \frac{( - 30 ) \times 15}{( - 30 + 15 )} = \frac{- 450}{- 15} = c m

From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).
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Question Text	A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?
Updated On	Sep 25, 2022
Topic	Ray Optics and Optical Instruments
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 4
Upvotes	392
Avg. Video Duration	8 min

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

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A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

Text solutionVerified

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