A light ray hits the pole of a thin biconvex | vedclass

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(C) At the pole of a thin lens,the surface is perpendicular to the optic axis. Therefore,the normal to the surface at the pole coincides with the opti

Vedclass · Accepted Answer

$(2/3)^o$

Vedclass · Answer

(C) At the pole of a thin lens,the surface is perpendicular to the optic axis. Therefore,the normal to the surface at the pole coincides with the optic axis.$1$. Applying Snell's Law at the first surface (pole):$\mu \sin(2^{\circ}) = (2\mu) \sin(r_1)$For small angles,$\sin(	heta) \approx 	heta$ (in radians),so:$\mu(2^{\circ}) = 2\mu(r_1) \implies r_1 = 1^{\circ}$$2$. Applying Snell's Law at the second surface:Since the lens is thin,the ray travels straight to the second surface at the same height (the pole). The normal at the second pole also coincides with the optic axis.$(2\mu) \sin(r_1) = (3\mu) \sin(e)$Using the small angle approximation:$2\mu(r_1) = 3\mu(e)$$2(1^{\circ}) = 3(e)$$e = (2/3)^{\circ}$Thus,the angle made by the emergent ray with the optic axis is $(2/3)^{\circ}$.

$A$ light ray hits the pole of a thin biconvex lens as shown in the figure. The angle made by the emergent ray with the optic axis will be

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