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A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change inits focal length when it is immersed in water of refraction index 4/3. |
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Answer» Solution :Here, it is given that `N=1.5, f_("air") = 18 cm` and `n_(m) = 4/3` `therefore` Focal LENGTH of lens in WATER `f_("water") =((n-1)n_(m)f_("air"))/(n-n_(m)) = ((1.5 -1) xx (4/3) xx 18)/(1.5-4/3) = 72 cm` `therefore` Change in focal length `=f_(m) - f_("air") = 72 cm - 18 cm = 54 cm ` |
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