Question
Derive $D_m=A\left(n_{21}-1\right)$ for thin lens.
✓
Answer
→Refractive index of prism $n_{21}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \frac{A}{2}} \ldots(1)$
→But for thin prism, prism angle A is small and,
$\begin{aligned}\therefore & \sin \left(\frac{ A + D _m}{2}\right) \approx \frac{ A + D _m}{2} \\& \sin \left(\frac{ A }{2}\right) \approx \frac{ A }{2}
\end{aligned}$
→Substituting in equation (1),
$\begin{array}{l}\therefore \quad n_{21}=\frac{\frac{A+D_m}{2}}{\frac{A}{2}} \\\therefore \quad n_{21}=\frac{ A + D _m}{A} \\\therefore n_{21} \cdot A = A + D _m \\\therefore \quad D _m=n_{21} \cdot A - A \\\therefore \quad D _m= A \left(n_{21}-1\right) \\\end{array}$
→From this formula it can be said that thin prism (Thin prism means ' A ' will be smaller and hence $D _{ m }$ will be smaller, too.) does not deflect light much.
→But for thin prism, prism angle A is small and,
$\begin{aligned}\therefore & \sin \left(\frac{ A + D _m}{2}\right) \approx \frac{ A + D _m}{2} \\& \sin \left(\frac{ A }{2}\right) \approx \frac{ A }{2}
\end{aligned}$
→Substituting in equation (1),
$\begin{array}{l}\therefore \quad n_{21}=\frac{\frac{A+D_m}{2}}{\frac{A}{2}} \\\therefore \quad n_{21}=\frac{ A + D _m}{A} \\\therefore n_{21} \cdot A = A + D _m \\\therefore \quad D _m=n_{21} \cdot A - A \\\therefore \quad D _m= A \left(n_{21}-1\right) \\\end{array}$
→From this formula it can be said that thin prism (Thin prism means ' A ' will be smaller and hence $D _{ m }$ will be smaller, too.) does not deflect light much.
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