Question

In a glass slab of thickness ${}^{\prime }{l}^{\prime }$ and refractive index ${\mu }_1$, a cuboidal cavity of thickness 'm' is carved and filled with a liquid of refractive index ${\mu }_2({\mu }_1>{\mu }_2)$. It has been found that shift produced by the given slab is zero when an observer 'A' observes an object 'B', the ratio $l/m$ will be :
(consider the incident rays to be paraxial)

• A. $\frac{{\mu }_1-{\mu }_2}{{\mu }_2-1}$
• B. $\frac{{\mu }_1-{\mu }_2}{{\mu }_2({\mu }_1-1)}$
• C. $\frac{{\mu }_1-{\mu }_2}{{\mu }_1-1}$
• D. $\frac{{\mu }_1-{\mu }_2}{{\mu }_1({\mu }_2-1)}$

Answer 1

The correct option is B $\frac{{\mu }_1-{\mu }_2}{{\mu }_2({\mu }_1-1)}$

When observer 'A' observes 'B', it can be visualized that the light ray will travel through two different media i.e. glass slab and cavity filled with liquid.
$\Rightarrow$ Thickness of cavity $\Rightarrow t_2=m$
Thickness of glass slab through which light passes, $t_1=(l-m)$
Net shift will be produced due to sum of shift produed by individual mediums:
$\Rightarrow S_{net}=S_1+S_2$
or, $S_{net}=t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})$
$(\because S_{net}=0)$
$\Rightarrow 0=(l-m)(1-\frac{1}{{\mu }_1})+m(1-\frac{1}{{\mu }_2})$
or, $0=(l-m)\left(\frac{{\mu }_1-1}{{\mu }_1}\right)+m\left(\frac{{\mu }_2-1}{{\mu }_2}\right)$
or, $-\frac{(l-m)}{m}\left(\frac{{\mu }_1-1}{{\mu }_1}\right)=\left(\frac{{\mu }_2-1}{{\mu }_2}\right)$
or, $(\frac{-l}{m}+1)=\frac{{\mu }_1({\mu }_2-1)}{{\mu }_2({\mu }_1-1)}$
or, $\frac{l}{m}=1-\frac{{\mu }_1({\mu }_2-1)}{{\mu }_2({\mu }_1-1)}$
or, $\frac{l}{m}=\frac{{\mu }_2({\mu }_1-1)-{\mu }_1({\mu }_2-1)}{{\mu }_2({\mu }_1-1)}$
$\therefore \frac{l}{m}=\frac{{\mu }_1-{\mu }_2}{{\mu }_2({\mu }_1-1)}$

Why this question? Tip: "Along the line of view 'AB'," the light ray will be effectively travelling through glass medium of thickness $(l-m)$ and liquid medium of thickness 'm'. It intends to challenge your visualization of path travelled by light ray.