Question

An object 4cm in sizeis placed 25cm in front of a concave mirror of focal length 15cm.At what distance from the mirror should a screen be placed in order to obtain a sharp image
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Answer 1

Given

• Object size (ho) = + 4cm
• Object distance (u) = - 25 cm
• Focal length (f) = - 15 cm

Find out

• Image distance (v) = ?

Solution

★Apply Mirror Formula★

$\implies\tt \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\tt \dfrac{1}{v}+\dfrac{-1}{25}=\dfrac{-1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{1}{25}-\dfrac{1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{3-5}{75} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{-2}{75} \\ \\ \\ \implies\tt v=\cancel\dfrac{-75}{2}\\ \\ \\ \implies\tt v=-37.5cm$

★Hence, the image distance of object is 37.5cm

$\rule{200}3$

Additional Information

• Magnification

It is the ratio of height or size of image to the height or size of object.

• m = hi/ho = -v/u

$\rule{200}3$

Answer 2

GIVEN :

Given, mirror is concave,

•Size of object (Ho) = +4cm.

•Object distance (u) = -25cm.

•Focal length of concave mirror (f) = - 15cm.

TO FIND :

•Image distance (v) = ?

SOLUTION :

=> Using mirror formula,

=> 1 / v + 1 / u = 1 / f

=> 1 / v + 1 /-25 = 1 / -15

=> 1 / v = 1 / -15 + 1 / 25

=> 1 / v = (-25 + 15) / 375

=> 1 / v = - 10 / 375

=> v = - 37.5 cm .

Therefore, the image distance of object is -37.5cm.