In a YDSE experiment, the two slits are covered with a transparent membrane of negligible thickness which allows light to pass through it but does not allow water. A glass slab of thickness t $$=$$ $$0.41$$ mm and refractive index μ$$g=1.5$$ is placed in front of one of the slits as shown in the figure. The separation between the slits is d $$=$$ $$0.30$$ mm. The entire space to the left of the slits is filled with water of refractive index μ$$w=4/3$$ .A coherent light of intensity I and absolute wavelength λ$$=5000Ao$$ is being incident on the slits making an angle $$30$$º with horizontal. If screen is placed at a distance D $$=1m$$ from the slits, then the distance of central maxima from O is n×$$0.83$$ cm. Then the value of n is

Question

In a YDSE experiment, the two slits are covered with a transparent membrane of negligible thickness which allows light to pass through it but does not allow water. A glass slab of thickness t $$=$$ $$0.41$$ mm and refractive index μ$$g=1.5$$ is placed in front of one of the slits as shown in the figure. The separation between the slits is d $$=$$ $$0.30$$ mm. The entire space to the left of the slits is filled with water of refractive index μ$$w=4/3$$ .A coherent light of intensity I and absolute wavelength λ$$=5000Ao$$ is being incident on the slits making an angle $$30$$º with horizontal. If screen is placed at a distance D $$=1m$$ from the slits, then the distance of central maxima from O is n×$$0.83$$ cm. Then the value of n is

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If central maxima is formed at point P on the screen, then the path difference due to water medium is cancelled by path difference due to glass slab.

$\begin{array}{l} For central bright spot, Δ x = 0 \\ \Rightarrow [S_{2} P + (μ_{g} - 1) t] - [S_{1} P + μ_{w} d \sin 30^{\circ}] = 0 \\ \Rightarrow S_{2} P - S_{1} P = \frac{x d}{D} = μ_{w} d \sin 30^{0} - (μ_{g} - 1) \times t \\ \Rightarrow x = - 1.66 cm \end{array}$

Coherent sources and their importance in interference

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Coherent sources and their importance in interference

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If central maxima is formed at point P on the screen, then the path difference due to water medium is cancelled by path difference due to glass slab. For central bright spot, Δx=0⇒S2P+μg−1t−S1P+μwdsin⁡30∘=0⇒S2P−S1P=xdD=μwdsin⁡300−μg−1×t⇒x=−1.66cm

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