Question

29. For single slit Diffraction, the path difference between the two ends of the slitis
(A) ∆=aSine
(B) ∆ = aCoso
(C) ∆ -Cose
(D) ∆=0​

Answer 1

Answer:

option a is correct option

Answer 2

Answer:

For single slit Diffraction, the path difference between the two ends of the slits is $\Delta=\mathrm{a} \sin \theta$ i.e.option (A).

Explanation:

• When monochromatic light is made incident on a single slit, there is diffraction of light at that slit. In front of the slit, on a screen, we observe diffraction patterns. Bands of light and darkness can be seen in the diffraction pattern. The intensity of the middle band is at its highest and continues to decline on either side.

• Assume that $\mathrm{AB}$ is a slit of width "a," on which a parallel monochromatic light beam is incident. According to Fresnel, a vast number of waves that originate from various lighted slit points are superimposed to create the diffraction pattern.
• Let $\theta$ be the angle of diffraction for waves reaching at point $P$ of the screen and $A N$  the perpendicular dropped from $\mathrm{A}$ on wave diffracted from $\mathrm{B}$. The path difference between rays diffracted at points $\mathrm{A}$ and $\mathrm{B}$.

                                           $\Delta=\mathrm{BP}-\mathrm{AP}=\mathrm{BN}$

In $\triangle \mathrm{ANB}$, $\angle \mathrm{ANB}=90^{\circ}$ and $\angle \mathrm{BAN}=\theta$

$\therefore \quad \sin \theta=\frac{\mathrm{BN}}{\mathrm{AB}} \text { or } \mathrm{BN}=\mathrm{AB} \sin \theta$

As $\mathrm{AB}=$ the Width of the slit $=\mathrm{a}$

$\therefore$ Path difference, $\Delta=\mathrm{a} \sin \theta$

Hence, for single slit Diffraction, the path difference between the two ends of the slits is $\Delta=\mathrm{a} \sin \theta$ i.e.option (A).

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