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Q. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

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A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

Solution:

By lens maker’s formula

$ frac{1}{f} = (mu_g – 1) bigg(frac{1}{R_1}-frac{1}{R_2}bigg) $ …(i)

When convex lens is dipped in a liquid of refractive index $ (mu_l ) $ then its focal length,

$ frac{1}{f_1} = bigg(frac{mu_g}{mu_l}-1bigg) bigg(frac{1}{R_1}-frac{1}{R_2}bigg) $

or $ frac{1}{f_1} = frac{(mu_g – mu_2)}{mu_l}bigg(frac{1}{R_1}-frac{1}{R_2}bigg) $ …(ii)

Dividing Eq. (i) by Eq. (ii), we get

$ frac{f_1}{f} = frac{(mu_g – 1)mu_l}{(mu_g -mu_l)} $ …(iii)

But it is given that refractive index of lens is equal to refractive index of liquid ie, $ mu_g =mu_l . $

Hence, Eq. (iii) gives,

$ frac{f_1}{f} = frac{(mu_g – 1)mu_l}{0}= infty (infinity) $

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