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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 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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प्रश्न

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.

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उत्तर

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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पाठ 9: Ray Optics and Optical Instruments - EXERCISES [पृष्ठ २४९]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.2 | पृष्ठ २४९
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
Exercise | Q 9.2 | पृष्ठ ३४४

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