Question

One face of a $6\text{cm}$ thick rectangular glass plate is silvered. An object held $8\text{cm}$ in front of the unsilvered face forms an image $10\text{cm}$ behind the silvered face. Find the critical angle for glass-air interface.
[consider two refractions and one reflection]

• A. ${\sin }^{-1}\left(\frac{1}{3}\right)$
• B. ${\sin }^{-1}\left(\frac{2}{3}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{2}{3}\right)$


For first refraction at $\text{AB}$,

$\mu =\frac{h_{\text{app}}}{h_{\text{real}}}=\frac{P{I}_1}{PO}$

$\Rightarrow P{I}_1=8\mu$

For first reflection at $\text{CD}$, $I_1$ will act as object. So,

$Q{I}_1=P{I}_1+QP=8\mu +6$

Thus, $Q{I}_2=8\mu +6$

This will form behind the mirror $\text{CD}$.

For second refraction at $\text{AB}$,

$P{I}_3=\frac{P{I}_2}{\mu }$ $[\mu =\frac{d_{\text{real}}}{d_{app}}]$

$\Rightarrow PQ+Q{I}_3=\frac{PQ+Q{I}_2}{\mu }$

$\Rightarrow 6+10=\frac{6+(8\mu +6)}{\mu }$

$\Rightarrow 16\mu =8\mu +12$

$\Rightarrow 8\mu =12\Rightarrow \mu =\frac{3}{2}$

Now, critical angle for glass-air interface,

${\theta }_{\text{c}}={\sin }^{-1}\left(\frac{{\mu }_{\text{r}}}{{\mu }_{\text{d}}}\right)$

$\Rightarrow {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1}{3/2}\right)$

$\Rightarrow {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{2}{3}\right)$

Hence, option $\left(b\right)$ is correct.