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What happens to the index of refraction and focal length of a lens that is initially in air that is then placed in water? $\endgroup$ 1 Text Solution Solution : From lens maker's formula <br> `(1)/(f) = (mu - 1)((1)/(R_1)-(1)/(R_2))` <br> `:. f prop (1)/((mu - 1))` <br> As `.^w mu_g lt .^a mu_g` <br> `:. f_w gt f_a`, i.e., focal length of lens will increase. |