First slide

Prism

Question

The refracting angle of prism is ${(\frac{14}{11})}^{0}$ and refractive index is 1.5. A thin convex lens of focal length 20 cm is placed at distance 30 cm from the point source S as shown in figure. The position of image of point source S is

Difficult

A

$\frac{1}{3}$ cm above the X-axis
B

$\frac{1}{3}$ cm below the X-axis
C

$\frac{2}{3}$ cm below the X-axis
D

$\frac{2}{3}$ cm above the X-axis

Solution

$\begin{array}{l} ∵ δ = (μ - 1) A \\ = (1.5 - 1) \frac{14}{11} \times \frac{π}{180} = \frac{1}{2} \times \frac{14}{11} \times \frac{22}{7 \times 180} = \frac{1}{90} \\ ∴ \tan δ = δ = \frac{h}{15} \\ \Rightarrow \frac{1}{90} = \frac{h}{15} \\ ∴ h = \frac{1}{6} c m \end{array}$
S’ behaves as object for the thin convex lens. For the lens, $u = - 30 c m, f = + 20 c m$
$\begin{array}{l} ∴ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ ∵ \frac{1}{v} - \frac{1}{- 30} = \frac{1}{20} \\ O r \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} \\ ∴ v = 60 c m \\ ∴ h^{'} = - 2 \times \frac{1}{6} = \frac{- 1}{3} c m [∵ \frac{h^{'}}{h} = \frac{v}{u}] \end{array}$

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