Question

An object of 5.0cm size is placed at a distance of 20.0cm from a converging mirror of focal length 15.0cm

Answer 1

Complete Question

An object 5 cm in size is placed at a distance of 20 cm from the concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed to get a sharp image? Also find the height of the image.

Note

Converging mirror means Concave mirror.

Solution

• u = - 20
• f = - 15

Mirror Formula

→ 1/f = 1/v + 1/u

→ 1/(- 15) = 1/v + 1/(- 20)

→ - 1/15 + 1/20 = 1/v

→ (- 4 + 3)/60 = 1/v

→ v = - 60 cm

Magnification

m = - v/u = h of image/h of object

→ - (- 60)/(- 20) = h/5

→ - 3 × 5 = h

→ - 15 = height of image

Answer

Screen should be placed 60 cm before mirror and height of image is 15 cm below principal axis.

Answer 2

• h' (height of object) = 5 cm

• u (object distance) = - 20.0 cm = - 20 cm

• f (focal length) = - 15 cm

_________ [GIVEN]

• We have to find the distance of the image from mirror (v) and height of the image (h).

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We know that..

$\boxed{ *\: \dfrac{1}{f} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{u}}$ (Mirror Formula)

Put the known values in above equation

=> $\dfrac{1}{-15} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{-20}$

=> $\dfrac{-1}{15}$ + $\dfrac{(-1)}{20}$ = $\dfrac{1}{v}$

=> $\dfrac{1}{v}$ = $\dfrac{-4\:+\:3}{60}$

=> $\dfrac{1}{v}$ = $\dfrac{1}{60}$

=> $\underline{v \:=\: 60\: cm}$

_______________________________

Now..

$\boxed{*\:m\:=\:\dfrac{-v}{u}\:=\:\dfrac{h}{h'}}$

=> $\dfrac{-60}{20}$ = $\dfrac{h}{5}$

=> $\dfrac{h}{5}$ = - 3

=> $\underline{h \:=\: - 15 \:cm}$

_________________ [ANSWER]

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