The 45^{circ}-45^{circ}-90^{circ} prism of he | vedclass

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(C) $1$. The light from point $P$ enters the prism. The apparent depth of $P$ as seen from inside the prism is $h' = \mu h = 2h$. The total distance o

Vedclass · Accepted Answer

$10$

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(C) $1$. The light from point $P$ enters the prism. The apparent depth of $P$ as seen from inside the prism is $h' = \mu h = 2h$. The total distance of the object from the silvered surface along the vertical path is $d_1 = 2h + x$,where $x$ is the distance from the top surface to the point on the hypotenuse.$2$. The silvered hypotenuse acts as a plane mirror. The image $I$ is formed at a distance $d_1 = 2h + x$ behind the mirror.$3$. This image $I$ acts as an object for the refraction at the vertical face of the prism. The distance of $I$ from the vertical face is $d_1 + y = 2h + x + y$,where $y$ is the distance from the point on the hypotenuse to the vertical face. From the geometry of the $45^{\circ}-45^{\circ}-90^{\circ}$ prism,$x + y = 10 \,cm$.$4$. The apparent distance of the image $I$ as seen from outside the prism is $d_{out} = \frac{d_1 + y}{\mu} = \frac{2h + 10}{2} = h + 5$.$5$. This image is at a distance $u = -(h + 5 + 15) = -(h + 20) \,cm$ from the convex lens. The lens forms an image on the wall at $v = +15 \,cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{15} - \frac{1}{-(h + 20)} = \frac{1}{10}$$\frac{1}{h + 20} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$$h + 20 = 30 \implies h = 10 \,cm$.

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