Explain with reason how the power of a diverging lens changes when (i) it is kept in a medium of refractive by index greater than that of the lens. (ii) incident red light is replaced by violet light.
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(i) Using standard notations, power of lens in air `P_(a) = (1)/(f_(a))=((mu_(g))/(mu_(a))-1)((1)/(R_(1)) - (1)/(R_(2)))` when the lens is immersed in a medium `P_(m) = (1)/(f_(m)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` As lens is diverging, `P_(a)` is negative. As `mu_(m) gtmu_(g)`, therefore `P_(m)` becomes positive. Hence power of the same lens immersed in medium becomes positive, i.e., diverging lens behave as converging lens of smaller power. (ii) When incident red light is replaced by violet light, wavelength decreases. Refractive index increases `(mu_(v) gt mu_(r ))`. Power of the lens increases.
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