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Karnataka Board PUCPUC Science 2nd PUC Class 12

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.

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Question

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

Numerical
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Solution

Height of the needle, h1 = 4.5 cm

Object distance, u = −12 cm

Focal length of the convex mirror, f = 15 cm

Image distance = v

The value of v can be obtained using the mirror formula:

`1/"u" + 1/"v" = 1/"f"`

`1/"v" = 1/"f" - 1/"u"`

`1/"v" = 1/15 - 1/-12`

`1/"v" = 1/15 + 1/12`

`1/"v" = (4 + 5)/60`

`1/"v" = 9/60`

v = `60/9`

∴ v = 6.67 cm

Hence, the image of the needle is 6.67 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" = "h"_2/"h"_1 = -"v"/"u"`

∴ `"h"_2 = - "v"/"u" xx "h"_1`

= `-6.67/(-12) xx 4.5`

= + 2.5 cm

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

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Chapter 9: Ray Optics and Optical Instruments - EXERCISES [Page 249]

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NCERT Physics Part I and II [English] Class 12
Chapter 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.2 | Page 249
NCERT Physics Part I and II [English] Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.2 | Page 344

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