Question

An object is placed at $8 cm$ from the upper face of a glass slab of thickness $6 cm$. 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. $6 cm$ behind the silvered face
• B. $8 cm$ behind the silvered face
• C. $10 cm$ behind the silvered face
• D. $4 cm$ behind the silvered face

Answer 1

Answer: (C)

For refraction at the first surface ab of the glass slab,
$\frac{\mu}{v}-\frac{1}{u}=\frac{\mu-1}{R}$
$\frac{\mu}{v}-\frac{1}{-8}=\frac{\mu-1}{\infty}$
$v =-8 \mu$
$v =-8 \times 1.5=-12 cm$
The image $I_{1}(v=-12 cm )$
will serve as an object for the plane mirror $cd$ and its
virtual image is formed at $I_{2}$ at a distance $u_{2}\left[u_{2}=-(12\right.$ $+6)=-18 cm ]$ behind the mirror. This image $I_{2}$ will serve as an object for the refracting surface ab for which $v=-(18+6)=-24 cm .$
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 $v=\frac{18}{\mu}=\frac{-24}{1.5}=-16 cm$
Thus, the final image I is formed at a distance $16 cm$ from the refracting face ab which means at a distance $16-6=10 cm$ behind the silvered face $cd$.