There are two isosceles prisms of prism angle $= {(\frac{1.8}{π})}^{\circ}$ and refractive index $1.5$ which are placed with their bases touching each other. This system can act as a converging lens on which parallel rays are incident having distance $1 cm$ from the line shown in figure. What will be distance of $F$ from base focal length (in meter) on the system ?

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Solution

Angle of deviation $δ = A (μ - 1) = \frac{1.8}{π} \frac{π}{180} (1.5 - 1) = \frac{1}{200}$
For small angles we know that $tan δ = δ$
$\frac{1 cm}{f} = δ$
$\Rightarrow f = 200 cm$
$\Rightarrow f = 2 m$

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Angle of deviation δ=A(μ−1)=1.8ππ180(1.5−1)=1200 For small angles we know that tanδ=δ 1 cmf=δ ⇒f=200 cm ⇒f=2 m

Angle of deviation $δ = A (μ - 1) = \frac{1.8}{π} \frac{π}{180} (1.5 - 1) = \frac{1}{200}$
For small angles we know that $tan δ = δ$
$\frac{1 cm}{f} = δ$
$\Rightarrow f = 200 cm$
$\Rightarrow f = 2 m$