In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. [2025]

Question

In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $d$ = 2 mm and the shortest distance between the slits and the screen is D = 2 m. Thickness of the combination of the wedges is $t = 12 μ m$ . The value of $l$ as shown in the figure is 1 mm. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to O by ________. [2025]

Smart Achievers · Accepted Answer

(1.2)

From figure, $t = 12 μ m$

$\frac{4}{12} = \frac{1}{x}$ $∴$ $x = 3 μ m, y = 9 μ m$

$∴$ $Δ = - \frac{y d}{D} - (μ_{B} - 1) x - (μ_{A} - 1) y + (μ_{B} - 1) y + (μ_{A} - 1) x$

$- \frac{y d}{D} = - 0.5 \times 3 - 0.7 \times 9 + 0.5 \times 9 + 0.7 \times 3$

$- \frac{y d}{D} = - 0.5 \times 6 - 0.7 \times 6$ $\Rightarrow - \frac{y d}{D} = - 1.2 μ m$

$\Rightarrow y = \frac{1.2 \times D}{d} \times 10^{- 6}$

$= \frac{1.2 \times 2}{2 \times 10^{- 3}} \times 10^{- 6}$

$= 1.2 mm$

Solution

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Solution

Ans. (1.2) From figure, x+y=12 μm 412=1x ∴ x=3 μm, y=9 μm ∴ Δ=-ydD-(μB-1)x-(μA-1)y+(μB-1)y+(μA-1)x -ydD=-0.5×3-0.7×9+0.5×9+0.7×3 -ydD=-0.5×6-0.7×6⇒-ydD=-1.2 μm ⇒y=1.2×Dd×10-6 =1.2×22×10-3×10-6 =1.2 mm

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