Consider the situation shown in the figure. Two slits S₁ and S₂ are placed symmetrically about the line OP which is perpendicular to screen. The space between screen and slits is filled with a liquid of refractive index $μ_{3}$ . A plate of thickness t and refractive index $μ_{2}$ is placed in front of one of the slit. A source S is placed above OP at a distance d. Choose the correct alternatives.
$[(Given that D = 1 m, d = 2 mm, t = 6 \times 10^{- 6} m, μ_{2} = 1 .2, μ_{3} = 1 .8]$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Position of central maxima from point P is 2 mm

If glass slab is removed, the central maxima shifts by a distance of 2 mm

Position of central maxima from point P is $\frac{2}{9} mm$

Thickness of glass slab so that central maxima forms at point P is $\frac{20}{3} \times 10^{- 6} m$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Optical path difference at P,

$\begin{array}{l} Δx = [{SS}_{2} + (S_{2} P - t) μ_{3} + μ_{2} t] - [{SS}_{1} + (S_{1} P) μ_{3}] \\ = (S_{2} - {SS}_{1}) + (S_{1} Q - S_{2} Q) μ_{3} + (μ_{2} - μ_{3}) t \\ = \frac{d^{2}}{D} + [μ_{3} \frac{yd}{D} + (μ_{2} - μ_{3}) t] \\ Δx = \frac{d^{2}}{D} + (μ_{2} - μ_{3}) t + \frac{μ_{3} yd}{2 D} \end{array}$