A screen is at a distance D = 80cm from a diaphragm having two...

A screen is at a distance $D = 80cm$ from a diaphragm having two narrow slits ${S}_{1}$ and ${S}_{2}$ which are $d = 2mm$ apart. The slit ${S}_{1}$ is covered by a transparent sheet of thickness ${t}_{1} = 2.5\mu m$ and the slit ${S}_{2}$ by another sheet of thickness ${t}_{2} = 1.25\mu m$ as shown in the figure. Both sheets are made of the same material having a refractive index ${μ}_{m} = 1.40$ . Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength $\lambda = 5000\text{Å}$ is incident normally on the diaphragm. Assuming the intensity of the beam to be uniform and slits of equal width, calculate the ratio of intensity at $C$ to the maximum intensity of interference pattern obtained on the screen, where $C$ is the foot of the perpendicular bisector of ${S}_{1}{S}_{2}$ . [Refractive index of water, ${μ}_{w} = \dfrac{4}{3}$ ]

A screen is at a distance D = 80cm from a diaphragm having two narrow slits S _ 1 and S _ 2 which are d = 2mm apart.

Options

A $\dfrac{3}{4}$
B $\dfrac{1}{2}$
C $\dfrac{1}{3}$
D $\dfrac{3}{5}$

Correct answer

A. $\dfrac{3}{4}$

Step-by-step solution

Optical path = (Refractive Index)

\times

(Geometrical Path Length)

Path Difference at point C on the screen

\Delta \text{x} = {\mu}_{\text{m}} {\text{t}}_{1} + {\mu}_{\text{w}} \left({\text{S}}_{1} \text{C} - {\text{t}}_{1}\right) - \left[{\mu}_{\text{m}} {\text{t}}_{2} + {\mu}_{\text{w}} \left({\text{S}}_{2} \text{C} - {\text{t}}_{2}\right)\right]

\Delta \text{x} = {\mu}_{\text{m}} \left({\text{t}}_{1} - {\text{t}}_{2}\right) - {\mu}_{\text{w}} {\text{t}}_{1} + {\mu}_{\text{w}} {\text{t}}_{2}

( ∵ {s}_{1} c = {s}_{2}c )

\Delta \text{x} = {\mu}_{\text{m}} \left({\text{t}}_{1} - {\text{t}}_{2}\right) - {\mu}_{\text{w}} \left({\text{t}}_{1} - {\text{t}}_{2}\right)

\Delta \text{x} = \left({\mu}_{\text{m}} - {\mu}_{\text{w}}\right) \left({\text{t}}_{1} - {\text{t}}_{2}\right)

\Delta \text{x} = \left(\text{1.4} - \dfrac{4}{3}\right) \left(\text{2.5} \times {\text{10}}^{- 6} - \text{1.25} \times {\text{10}}^{- 6}\right)

\Delta \text{x} = \dfrac{\text{0.2}}{3} \times \text{1.25} \times {\text{10}}^{- 6}

\Delta \text{x} = \dfrac{\text{2.5}}{3} \times 1 {0}^{- 7} \text{m}

\Delta \text{x} = \dfrac{\text{2500}}{3} Å = \dfrac{\text{5000}}{6} Å

\Delta \text{x} = \dfrac{\lambda}{6} \text{,}

Phase difference

ϕ = \dfrac{\pi}{3}

= {60}^{ο}

at any point on the screen

{I}_{net} = {I}_{1} + {I}_{2} + 2\sqrt{{I}_{1} {I}_{2}} cosϕ

Resultant intensity

{\text{I}}_{\text{C}} = \text{I} + \text{I} + 2 \sqrt{\text{I} \times \text{I}} \text{cos} 6 0

{\text{I}}_{\text{C}} = 3\text{I}

{\text{I}}_{\text{max}} = \text{I} + \text{I} + 2\text{I}

when

ϕ = {0}^{o} ⇒ cosϕ = + 1

{\text{I}}_{\text{max}} = 4\text{I}

\dfrac{{\text{I}}_{\text{C}}}{{\text{I}}_{\text{max}}} = \dfrac{3}{4}

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