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Question: The distance between the slit and the bi-prism and that of between bi-prism and screen each is 0.4 m...

The distance between the slit and the bi-prism and that of between bi-prism and screen each is 0.4 m. The obtuse angle of bi-prism is 179∘179^\circ and the refractive index is 1.5. If the fringe width is 1.8×10−4m1.8 \times {10^{ - 4}}m then the distance between imaginary sources will be
(A) 8.7 mm
(B) 4.36 mm
(C)1.5 mm
(D) 3.5 mm

Explanation

Solution

Hint
Firstly, we will find out the acute angle of biprism in radians. Then we will find the relation between deviation of the source beam and the angle of the prism. We then find out that the total deviation can be calculated by ( =δ×= \delta \timesdistance between source and prism) this formula. Then we will find our required solutions after multiplying 2 with total deviation.

Complete step by step answer
Given that,
Obtuse angle in the bi prism is 179∘179^\circ
So, each acute angle of the prism is,
A=180∘−179∘2A = \dfrac{{180^\circ - 179^\circ }}{2}
⇒A=0.5∘\Rightarrow A = 0.5^\circ
⇒A=0.5×π180\Rightarrow A = \dfrac{{0.5 \times \pi }}{{180}} radians
We know that,
μ=sin⁡(A+δ2)sin⁡A2\mu = \dfrac{{\sin \left( {\dfrac{{A + \delta }}{2}} \right)}}{{\sin \dfrac{A}{2}}} [A = prism angle, μ\mu = refractive index, δ\delta = Deviation of the source beam]
⇒μ=A+δ2A2\Rightarrow \mu = \dfrac{{A + \dfrac{\delta }{2}}}{{\dfrac{A}{2}}}
μA=A+δ\mu A = A + \delta
A(μ−1)=δA(\mu - 1) = \delta
sin⁡θ=θ\sin \theta = \theta as θ\theta is very small angle
Deviation of the source beam,
=δ= \delta
=A(μ−1)= A(\mu - 1)
Total deviation of each beam,
=δ×= \delta \timesdistance between source and prism
=0.4×A(μ−1)= 0.4 \times A(\mu - 1)
Distance between imaginary sources, d = 2 ×\times total deviation of each beam
d=2×0.4×0.5×π180(1.5−1)d = 2 \times 0.4 \times \dfrac{{0.5 \times \pi }}{{180}}\left( {1.5 - 1} \right)
∴d=3.49×10−3m ⇒d≈3.5mm  \therefore d = 3.49 \times {10^{ - 3}}m \\\ \Rightarrow d \approx 3.5mm \\\
Hence the required solution is 3.5mm (Option-A).

Note
One may think that the fringe width is given in the question. There is given extra information about fringe width if we want to find out the wavelength of the beam. Distance between imaginary sources can easily be calculated after finding the deviation of the beam.