Question

A rectangular block is composed of three different glass prisms (with refractive indices ${\mu }_1,{\mu }_2$ and ${\mu }_3$) as shown in the figure below. A ray of light incident to the left face emerges normal to the right face. Then the refractive indices are related as:

• A. ${\mu }_1^2+{\mu }_2^2=2{\mu }_3^2$
• B. ${\mu }_1^2+{\mu }_2^2={\mu }_3^2$
• C. ${\mu }_1^2+{\mu }_3^2=2{\mu }_2^2$
• D. ${\mu }_2^2+{\mu }_3^2=2{\mu }_1^2$

Answer 1

The correct option is C ${\mu }_1^2+{\mu }_3^2=2{\mu }_2^2$

From the figure,

$i_1=r_2={45}^o$

In $\Delta AEF,$,

$\angle A+\angle E+\angle F={180}^o$

$⟹\angle A={90}^o$

In quadrilateral $ABCD,$

$\angle A+\angle B+\angle C+\angle D={180}^o$

$⟹\angle C={90}^o$

in $\Delta BCD,$

$\angle B+\angle C+\angle D={180}^o$

$r_1+i_2={90}^o$

$\sin r_1=\cos i_2$

${\sin }^2{r}_1={\cos }^2{i}_2$

${\sin }^2{r}_1=1-{\sin }^2{i}_2$

Using snell's law,

$\frac{{\sin }^2{i}_1}{({\mu }_2/{\mu }_1{)}^2}=1-({\mu }_3/{\mu }_2{)}^2{\sin }^2{r}_2$

$\frac{{\mu }_1^2}{2{\mu }_2^2}+\frac{{\mu }_3^2}{2{\mu }_2^2}=1$

${\mu }_1^2+{\mu }_3^2=2{\mu }_2^2$