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Chapter 18: Problem 104

Path difference for the first secondary maximum in the Fraunhofer diffraction pattern of a single slit is given by ( \(a\) is the width of the slit) (A) \(a \sin \theta=\frac{\lambda}{2}\) (B) \(a \cos \theta=\frac{3 \lambda}{2}\) (C) \(a \sin \theta=\lambda\) (D) \(a \sin \theta=\frac{3 \lambda}{2}\)

Short Answer

Expert verified
The correct expression for the path difference for the first secondary maximum in the Fraunhofer diffraction pattern of a single slit is \(a \sin \theta=\lambda\), which corresponds to option (C).

Step by step solution

01

Consider the Fraunhofer diffraction pattern

We know that the Fraunhofer diffraction pattern is generated when light undergoes constructive interference after passing through a single slit. The condition for constructive interference is given by the path difference being an integral multiple of the wavelength, i.e., \(m \lambda = a \sin \theta\), where m is an integer representing the order of interference, a is the width of the slit, and 胃 is the angle at which the light is diffracted. We need to find the path difference that corresponds to the first secondary maximum.
02

Determine the first secondary maximum value

The first secondary maximum occurs just after the central maximum. Since the central maximum corresponds to m = 0, the first secondary maximum corresponds to m = 1 (as we need to consider only positive integer values for m). Using the expression \(m \lambda = a \sin \theta\) for constructive interference, we can write the equation for the first secondary maximum as, \(1 * \lambda = a \sin \theta\), which simplifies to, \(a \sin \theta = \lambda\)
03

Compare the given options with the derived expression

Now, let's compare the derived expression (\(a \sin \theta = \lambda\)) with the given options: (A) \(a \sin \theta=\frac{\lambda}{2}\) (B) \(a \cos \theta=\frac{3 \lambda}{2}\) (C) \(a \sin \theta=\lambda\) (D) \(a \sin \theta=\frac{3 \lambda}{2}\) We can see that the expression we derived for the first secondary maximum matches with option (C): \(a \sin \theta=\lambda\)
04

Conclusion

After analyzing the Fraunhofer diffraction pattern and the principle of constructive interference, we determined that the correct expression for the path difference for the first secondary maximum is \(a \sin \theta=\lambda\), which corresponds to option (C).

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Most popular questions from this chapter

In YDSE distance between the slits plane and screen is \(1 \mathrm{~m}\) and distance between two slits is \(5 \mathrm{~mm}\). If slabs of thickness \(2 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) having refractive index \(1.5\) and \(1.4\) are placed in front of two slits, the shift of central maximum will be (A) \(2 \mathrm{~m}\) (B) \(8 \mathrm{~cm}\) (C) \(20 \mathrm{~cm}\) (D) \(80 \mathrm{~cm}\)

A monochromatic light of wavelength \(4500 \AA\) fall on a single slit of width \(1.5 \mathrm{~mm}\) and resulting diffraction pattern is observed with the help of a lens of focal length \(10 \mathrm{~m}\). Angular width of central bright fringe will be (A) \(6 \times 10^{-40}\) (B) \(0.034^{\circ}\) (C) \(0.034\) radian (D) \(2.4^{\circ}\)

A point object is placed at \(30 \mathrm{~cm}\) from a convex glass lens \(\left(\mu_{g}=\frac{3}{2}\right)\) of focal length \(20 \mathrm{~cm}\). The final image of object will be formed at infinity if (A) Another concave lens of focal length \(60 \mathrm{~cm}\) is placed in contact with the previous lens. (B) Another convex lens of focal length \(60 \mathrm{~cm}\) is placed at a distance of \(30 \mathrm{~cm}\) from the first lens. (C) The whole system is immersed in a liquid of refractive index \(\frac{4}{3}\). (D) The whole system is immersed in a liquid of refractive index \(\frac{9}{8}\)

We have two equilateral prisms \(A\) and \(B\). They are made of materials having refractive index \(1.5\) and \(1.6 .\) For minimum deviation, incident angle will be (A) Small for prism \(A\) (B) Small for prism \(B\) (C) Equal for both the prisms (D) Can't be predicted

A ray incident at a point at an angle of incidence of \(60^{\circ}\) enters a glass sphere of R.1. \(n=\sqrt{3}\) and gets reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is (A) \(50^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(40^{\circ}\)
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