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CBSE Class 10 Science 2016 Term II Outside Delhi Set 2

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Question : 12 of 12
Marks: +1, -0
(a) Define focal length of a spherical lens.
(b) A divergent lens has a focal length of 30 cm30\text{ cm}. At what distance should an object of height 5 cm5\text{ cm} from the optical centre of the lens be placed so that its image is formed 15 cm15\text{ cm} away from the lens? Find the size of the image also.
(c) Draw a ray diagram to show the formation of image in the above situation.
Solution:  
(a) The distance between the optical centre and focus of a spherical lens is called focal length.
(b) Given : f=−30 cm,hi=5 cm,hi=f=-30\text{ cm}, h_i=5\text{ cm}, h_i= ?,
v=−15 cm,u=   ?   v=-15\text{ cm}, u=\;\text{ ? }\;
We know that,
or
  1f  =  1v−  1u\;\frac{1}{f}\;=\;\frac{1}{v}-\;\frac{1}{u}
  1u  =  1v−  1f\;\frac{1}{u}\;=\;\frac{1}{v}-\;\frac{1}{f}
u  =  vff−vu\;=\;\frac{v f}{f-v}
  =  −15×−30−30+15=−30 cm.\;=\;\frac{-15 \times -30}{-30+15}=-30\text{ cm} .
⇒    u=  vff−v\Rightarrow \;\; u=\;\frac{v f}{f-v}
Now,
   Now,         m=  vu=  hih0\;\text{ Now, }\;\; \;\; m=\;\frac{v}{u}=\;\frac{h_i}{h_0}
⇒hi=  vu×h0=  −15−30×5=2.5 cm.\Rightarrow h_i=\;\frac{v}{u} \times h_0=\;\frac{-15}{-30} \times 5=2.5\text{ cm} .
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