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Question

In the given figure a plano-concave lens is placed on a paper on which flower is drawn. How far above its actual position does the flower appear to be ?
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A
10 cm
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B
15 cm
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C
50 cm
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D
none of these
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Solution

The correct option is A 10 cm
Given: In the given figure a plano-concave lens is placed on a paper on which flower is drawn. 
To find how far above its actual position does the flower appear
Solution:
Refraction aoccur at the surface, 
and according to the given criteria, 
object distance, $$u=-20cm$$
refractive index, $$\mu_1=\dfrac 32, \mu_2=1$$
And radius of curvature, $$R=+20cm$$
Applying the lens formula, we get
$$\dfrac {\mu_2}v-\dfrac {\mu_1}u=\dfrac {\mu_2-\mu_1}R\\\implies \dfrac 1v-\dfrac {\dfrac 32}{(-20)}=\dfrac {1-\dfrac 32}{20}\\\implies \dfrac 1v=\dfrac {3}{-40}+\dfrac {2-3}{40}\\\implies \dfrac 1v=\dfrac {-3-1}{40}\\\implies v=-10cm$$
Hence the flower will appear 10 cm above its actual position of flower.

Physics

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