Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given by $\mathrm{x}=\mathrm{n} \lambda(\mathrm{D} / \mathrm{d})$ and the position of dark fringes is given by \(\mathrm{x}=(2 \mathrm{n}-1)(N 2)(\mathrm{D} / \mathrm{d})\) where \(\mathrm{n}=1,2,3 \ldots \ldots \ldots \ldots\) for first second, third bright/dark fringe. The center of the fringe pattern is bright (for \(\mathrm{n}=0\) ). The width of each bright/dark fringe is $\beta=(\lambda \mathrm{D} / \mathrm{d})\(, Where \)\lambda=5000 \AA\(. Slits are \)0.2 \mathrm{~cm}$ apart and \(\mathrm{D}=1 \mathrm{~m}\) (i) If light of wavelength \(6000 \AA\) be used in the above experiment the fringe width would be \(\mathrm{mm}\) (A) \(0.30\) (B) 3 (C) \(0.6\) (D) 6 (ii) with the light of wavelength \(5000 \AA\), If experiment were carried out under water of a \(n=(4 / 3)\) the fringe width would be (A) zero (B) \((4 / 3)\) times (C) (3/4) times (D) none of these

Step by step solution

01

Question 1: Find the fringe width with λ = 6000 Å

First, we need to convert the wavelength from Å to meters: \(6000 ~\mathrm{Å} = 6000 \cdot 10^{-10} ~\mathrm{m} = 6 \cdot 10^{-7} ~\mathrm{m}\) Now, we can plug in the values into the formula for fringe width: \(\beta = \frac{\lambda D}{d} = \frac{6 \cdot 10^{-7} \cdot 1}{0.2 \cdot 10^{-2}}\) Evaluating the above expression: \(\beta = 3 \cdot 10^{-3} ~\mathrm{m} = 3 ~\mathrm{mm}\) Therefore, the fringe width with a light of wavelength 6000 Å is \(3 ~\mathrm{mm}\). The correct answer is (B).

02

Question 2: Find the fringe width with λ = 5000 Å underwater

First, we need to convert the wavelength from Å to meters and account for the refractive index of water (4/3): \(\lambda' = \frac{\lambda}{n} = \frac{5000 \cdot 10^{-10}}{(4/3)} = 3.75 \cdot 10^{-7} ~\mathrm{m}\) Now, plug in the new values into the formula for fringe width: \(\beta' = \frac{\lambda' D}{d} = \frac{3.75 \cdot 10^{-7} \cdot 1}{0.2 \cdot 10^{-2}}\) Evaluating the above expression: \(\beta' = 1.875 \cdot 10^{-3} ~\mathrm{m} = 1.875 ~\mathrm{mm}\) Since we want to find the ratio with respect to the original fringe width using light of wavelength 5000 Å, we can simply take the ratio of the original fringe width to the new fringe width. \(\frac{\beta'}{\beta} = \frac{1.875}{2.5} = \frac{3}{4}\) Therefore, the fringe width underwater with light of wavelength 5000 Å is (3/4) times the original fringe width. The correct answer is (C).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!