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Home » An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 
An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 
CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments
An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 

Answer:

According to the question,

Size of the object is h_{1} = 3 cm

Object distance is u = -14 cm

Focal length of the given concave lens is f = -21 cm

Image distance is denoted by v

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}

=\frac{-2-3}{42}=\frac{-5}{42}

Therefore\;v=-\frac{42}{5}=-8.4cm

As a result, the image obtained is 8.4cm away on the opposite side of the lens. Because there is a negative sign, the image is assumed to be virtual and upright.

m=\frac{Image\;height(h_{2})}{Object\;height(h_{1})}=\frac{v}{u}

Therefore\;h_{2}=\frac{-8.4}{-14}\times3=0.6\times3=1.8cm

The height of the image is 1.8cm.

The virtual image will travel towards the mirror when the object is moved further away from the lens. The image size reduces as the object distance rises.

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