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Chapter 18: Problem 175

A mixture of light, consisting of wavelength \(590 \mathrm{~nm}\) and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4 th bright fringe of the unknown light. From this data, the wavelength of the unknown light is (A) \(885.0 \mathrm{~nm}\) (B) \(442.5 \mathrm{~nm}\) (C) \(776.8 \mathrm{~nm}\) (D) \(393.4 \mathrm{~nm}\)

Short Answer

Expert verified
The unknown wavelength is \(442.5 \text{ nm}\), which corresponds to option (B).

Step by step solution

01

Identify the formula

We will use the formula for bright fringes in a double-slit interference pattern: \(m\lambda = d\sin\theta\), where m is the fringe order, 位 is the wavelength, d is the slit separation, and 胃 is the angle between the central maximum and the m-th bright fringe.
02

Find the relationship between known and unknown wavelengths

Since the 3rd bright fringe of known light coincides with the 4th bright fringe of the unknown light, we can write the following equations: For known light: \(3\lambda_{known} = d\sin\theta_{known}\) For unknown light: \(4\lambda_{unknown} = d\sin\theta_{unknown}\) We are given that the central maximum of both lights coincides, hence both fringe patterns have the same angle. Therefore, we can say that sin(胃_known) = sin(胃_unknown).
03

Eliminate d and solve for unknown wavelength

Since \(\sin\theta_{known} = \sin\theta_{unknown}\), we have: \(3\lambda_{known} = 4\lambda_{unknown}\). Now, plug in the known wavelength (590 nm) and solve for the unknown wavelength: \(3 (590 \text{ nm}) = 4\lambda_{unknown}\) Divide by 4 to find the unknown wavelength: \(\lambda_{unknown} = \frac{3 (590 \text{ nm})}{4} = 442.5 \text{ nm}\). So, the unknown wavelength is 442.5 nm, which corresponds to option (B).

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Most popular questions from this chapter

The maximum intensity in Young's double slit experiment is \(I_{0}\). Distance between the slits is \(d=5 \lambda\), where \(\lambda\) is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance \(D=10 d\) ? (A) \(\frac{I_{0}}{2}\) (B) \(\frac{3 I_{0}}{4}\) (C) \(I_{0}\) (D) \(\frac{I_{0}}{4}\)

A microscope is focused on a needle lying in an empty tank. Now, the tank is filled with benzene to a height \(120 \mathrm{~mm}\). The microscope is moved \(40 \mathrm{~mm}\) to focus the needle again. The refractive index of benzene is (A) \(1.5\) (B) \(2.5\) (C) \(3.0\) (D) \(4.5\)

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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

In a Young's double slit experiment, the slit separation is \(1 \mathrm{~mm}\) and the screen is \(1 \mathrm{~m}\) from the slit. For a monochromatic light of wavelength \(500 \mathrm{~nm}\), the distance of third minimum from the central maximum is (A) \(0.50 \mathrm{~mm}\) (B) \(1.25 \mathrm{~mm}\) (C) \(1.50 \mathrm{~mm}\) (D) \(1.75 \mathrm{~mm}\)
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