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Physics Part-II

Ray Optics and Optical Instruments

A 4.5 cm needle is placed 12 cm away from a convex mirror of f

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

A small candle,

2.5 c m

in size is placed at

27 c m

in front of a concave mirror of radius of curvature

36 c m .

At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

4.5 c m

needle is placed

12 c m

away from a convex mirror of focal length

15 c m

. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

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Question

Easy

Solving time: 3 mins

A $4.5 c m$ needle is placed $12 c m$ away from a convex mirror of focal length $15 c m$ . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

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Updated on: Jun 26, 2023

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Physics Part-II (NCERT)

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Oliver Discussed

4.5 c m

needle is placed

12 c m

away from a convex mirror of focal length

15 c m

. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Given: Height of the needle,

h_{1} = 4.5 c m

Object distance,

u = - 12 c m

The focal length of the convex mirror,

f = 15 c m

Image distance,

v

The value of

v

can be obtained using the mirror formula.

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} + \frac{1}{- 12} = \frac{1}{15}

\frac{1}{v} = \frac{1}{12} + \frac{1}{15}

\frac{1}{v} = \frac{9}{60}

∴ v \approx 6.7 c m

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.

The image size is given by the magnification formula.

m = \frac{h ^{^{'}}}{h} = - \frac{v}{u}

h^{'} = \frac{6.7 \times 4.5}{12}

\Rightarrow h^{'} = + 2.5 c m

So,

m = \frac{2.5}{4.5}

m = 0.56

The height of the image is

2.5 c m

. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.

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Practice more questions from Physics Part-II (NCERT)

A small candle,

2.5 c m

in size is placed at

27 c m

in front of a concave mirror of radius of curvature

36 c m .

4.5 c m

needle is placed

12 c m

away from a convex mirror of focal length

15 c m

. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

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Question Text	A $4.5 c m$ needle is placed $12 c m$ away from a convex mirror of focal length $15 c m$ . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Updated On	Jun 26, 2023
Topic	Ray Optics and Optical Instruments
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 9
Upvotes	1039
Avg. Video Duration	8 min

A $4.5 c m$ needle is placed $12 c m$ away from a convex mirror of focal length $15 c m$ . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Text solutionVerified

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A $4.5 c m$ needle is placed $12 c m$ away from a convex mirror of focal length $15 c m$ . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.