Question

An object is placed at $8cm$ from the upper face of a glass slab of thickness $6cm$. The lower face of the slab is silvered as shown in the figure. If the refractive index of the material of slab is $1.5,$ find the position of the image:

• A. $6cm$ behind the silvered face
• B. $8cm$ behind the silvered face
• C. $10cm$ behind the silvered face
• D. $4cm$ behind the silvered face

Answer 1

Answer: (C)

For refraction at the first surface ab of the glass slab,
$\frac{\mu }{v}-\frac{1}{u}=\frac{\mu -1}{R}$
$\frac{\mu }{v}-\frac{1}{-8}=\frac{\mu -1}{\infty }$
$v=-8\mu$
$v=-8\times 1.5=-12cm$
The image $I_1(v=-12cm)$
will serve as an object for the plane mirror $cd$ and its
virtual image is formed at $I_2$ at a distance $u_2[u_2=-(12$ $+6)=-18cm]$ behind the mirror. This image $I_2$ will serve as an object for the refracting surface ab for which $v=-(18+6)=-24cm.$
The final image $I$ is formed at a distance $v$ from the face ab.
$\frac{\frac{1}{\mu }}{v}=\frac{1}{u}=\frac{\frac{1}{\mu }-1}{R}$
or $\frac{1}{v}-\frac{\mu }{u}=\frac{1-\mu }{R}$
or $\frac{1}{v}-\frac{\mu }{-18}=\frac{1-\mu }{\infty }$
or $v=\frac{18}{\mu }=\frac{-24}{1.5}=-16cm$
Thus, the final image I is formed at a distance $16cm$ from the refracting face ab which means at a distance $16-6=10cm$ behind the silvered face $cd$.