A screen is at a distance D=80cm from a diaphragm having two narrow slits S₁ and S₂ which are d = 2mm apart. Slit S₁ is covered by a transparent sheet of thickness t₁ = 2.5 $μ$ m and slit 2 S is covered by another sheet of thickness t₂ =1.25 $μ$ m as shown in figure. Both sheets are made of same material having refractive index $μ = 1 .4 .$ . Water is filled in the space between the diaphragm and screen. A monochromatic light beam of wavelength $λ = 5000 A^{0}$ is incident normally on the diaphragm. Assuming intensity of beam to be uniform calculate ratio of intensity at C to the maximum intensity of interference pattern obtained on the screen? $[μ_{ω} = 4 / 3]$

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$\frac{3}{4}$

$\frac{2}{3}$

$\frac{8}{9}$

$\frac{5}{7}$

answer is A.

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Detailed Solution

Path difference at ‘C”
$\begin{array}{l} Δx = t_{1} (_{ω} μ_{g} - 1) - t_{2} (_{ω} μ_{g} - 1) =_{ω} μ_{g} (t_{1} - t_{2}) - (t_{1} - t_{2}) \\ Δx = (t_{1} - t_{2}) (_{ω} μ_{g} - 1) \\ Δx = (t_{1} - t_{2}) (\frac{μ_{g}}{μ_{ω}} - 1) = (2 .5 - 1 .25) [\frac{1 .4}{4} \times 3 - 1] \\ = 1 .25 \times [\frac{42}{40} - 1] = 1 .25 \times \frac{2}{40} = \frac{2 .5}{40} \\ Δx = \frac{25}{400} = \frac{1}{16} μm . \end{array}$
Phase difference $(ϕ) = \frac{2 π}{λ_{ω}} \times Δx$
$ϕ = \frac{2 π}{μ_{a} λ_{a}} \times μ_{ω} \times Δx = \frac{2 π \times 4}{1 \times 5000 \times 10^{- 10} \times 3} \times \frac{1}{16} \times 10^{- 6}$
$ϕ = \frac{π \times 10^{- 6}}{30 \times 10^{- 7}} = \frac{π \times 10^{- 6}}{3 \times 10^{- 6}} = \frac{π}{3}$
intensity at ‘C’
$\begin{array}{l} I_{C} = 4 I_{0} \cos^{2} (ϕ / 2) = 4 I_{0} \cos^{2} (π / 6) = 4 I_{0} \times \frac{3}{4} \\ I_{C} = 3 I_{0} \\ maximum intensity I_{max} = 4 I_{0} \\ \frac{I_{C}}{I_{max}} = \frac{3 I_{0}}{4 I_{0}} \Rightarrow \frac{I_{C}}{I_{max}} = \frac{3}{4} \end{array}$