Fewpal
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A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

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According to the question; 

Object distance = u; 

Image distance (v) = -15cm; 

Focal length = -20cm 

By lens formula;

\(\frac1v-\frac1u=\frac1f\)

⇒ \(\frac1{15}-\frac1u=\frac1{-20}\)

⇒ \(\frac1{u}=\frac1{20}-\frac1{15}\)

⇒ \(\frac1{u}=\frac{3-4}{60}=\frac{-1}{60}\)

⇒ u = -60cm. 

Therefore, object is placed at 60 cm in front of lens. 

Now; 

Height of object h1= 5cm;

Magnification = \(\frac{h_2}{h_1}=\frac{v}{u}\)

Putting values of v and u

Magnification = \(\frac{h_2}{5}=\frac{-15}{-60}\)

⇒ \(\frac{h_2}{5}=\frac{1}{4};\)

⇒ h2\(\frac54\) = 1.25

Height of image is 1.25 cm. 

Positive sign means image is virtual.

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