(a) at $P_2$

(b) at $\large\frac{3R}{2}$ from $P_2$ towards left

(c)at $\large\frac{5R}{2}$ from $P_2$ towards right

(d)at $\large\frac{5R}{2}$ from $P_2$ towards left $

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(a) at $P_2$

(b) at $\large\frac{3R}{2}$ from $P_2$ towards left

(c)at $\large\frac{5R}{2}$ from $P_2$ towards right

(d)at $\large\frac{5R}{2}$ from $P_2$ towards left $

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Refraction at first surface

$u=\infty; \mu _1=\frac{4}{3} ;\mu_2 =1 ;Radius of curvature=+R$

Using equation $\large\frac{1}{v} -\frac{4/3}{( -\infty)}=\frac{1-\Large\frac{4}{3}}{+R}$

$V=-3R$

The image is at distance 3R from $P_1$ in water .

Refraction at second surface.

$\mu_2=\large\frac{4}{3}$$; \mu_1=1$

$u= -(3R+2R) =-5 R ;Radius of curvature=-R$

$\therefore \large\frac{\Large\frac{4}{3}}{v}-\frac{1}{-5R}=\frac{\Large\frac{4}{3}-1}{-R}$

The final image is at a distance $\large\frac{5R}{2}$ from $P_2$ towards left.

Hence d is the correct answer.

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