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  • 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?

Q 9.9  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?

Answers (1)

best_answer

In any Lens 

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

\\v= the distance of the image from the optical centre

\\u= the distance of the object from the optical centre

\\f= the focal length of the lens

Here Given,

\\u= -14 cm

\\f= -21 cm

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

\frac{1}{v} =\frac{1}{-21}- \frac{1}{14} = \frac{-5}{42}

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

Hence image distance is -8.4 cm. the negative sign indicates the image is erect and virtual.

Also as we know,]

m= -\frac{v}{u} = \frac{h'}{h}

From Here 

h'= -\frac{v}{u}h

h'= -\frac{-8.4}{-12}*3= 1.8 cm

Hence the height of the image is 1.8 cm.

As we move object further away from the lens, the image will shift toward the focus of the lens but will never go beyond that.size of the object will decrease as we move away from the lens.

Posted by

Pankaj Sanodiya

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