The relation between n_1 and n_2,if the behav | vedclass

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(B) The lens shown in the figure is a convex lens. In air,a convex lens is converging. However,the figure shows that the parallel light rays are diver

Vedclass · Accepted Answer

$n_2 > n_1$

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(B) The lens shown in the figure is a convex lens. In air,a convex lens is converging. However,the figure shows that the parallel light rays are diverging after passing through the lens.According to the Lens Maker's Formula,the focal length $f$ of a lens of refractive index $n_1$ placed in a medium of refractive index $n_2$ is given by:$\frac{1}{f} = (\frac{n_1}{n_2} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$For a convex lens,$(1/R_1 - 1/R_2) > 0$. If the lens behaves as a diverging lens (rays diverge),then $f This implies that $(\frac{n_1}{n_2} - 1)  n_1$.Therefore,the refractive index of the surrounding medium $n_2$ must be greater than the refractive index of the lens material $n_1$.

The relation between $n_1$ and $n_2$,if the behavior of light rays is as shown in the figure,is:

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