1.

A convex lens of focal legnth `0.2 m` and made of glass `(mu = 1.50)` is immersed in water `(mu = 1.33)`. Find the change in the focal length of the lens.A. `5.8 m`B. `0.58 cm`C. `0.58m`D. `5.8 cm`

Answer» Correct Answer - C
Using `(1)/(f_(a)) = (.^(a)mu_(g) - 1) ((1)/(R_(1))- (1)/(R_(2)))`
Here, `f_(a) = 0.2 m, .^(a)mu_(g) = 1.50`
`:. (1)/(0.2) = (1.50 - 1) ((1)/(R_(1)) - 1/(R_(2))) rArr (1)/(R_(1)) - (1)/(R_(2)) = 10`
Consider `f_(w)` be the focal length of the lens , when immersed in water.
`.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.50)/(1.33) = 1.128`
Now, `(1)/(f_(w)) = (.^(w)mu_(g) - 1)((1)/(R_(1)) - (1)/(R_(2))) = (1.28 - 1) xx 10 = 1.28`
or `f_(w) = (1)/(1.28) = 0.78`
Hence, change in forcal length of the lens is
`f_(w) - f_(a) = 0.78 - 0.2 = 0.58 m`


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