SATHEE: wave-optics Question 16

wave-optics Question 16

Question: Q. 5. Laser light of wavelength $640 nm$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $7.2 mm$ . Calculate the wavelength of another source of light which produces interference fringes separated by $8.1 mm$ using same arrangement. Also find the minimum value of the order $(n)$ of bright fringe of shorter wavelength which coincides with that of the longer wavelength. A [O.D. Comptt. I, II, III 2012]

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Solution:

Ans. Fringe width in interference pattern

$\begin{matrix} (i) & β = \frac{D λ}{d} \end{matrix}$

According to the question when $λ = 640 nm$ then $β = 7.2 mm$ . Putting these values in the relation (i)

$\begin{aligned} 7.2 \times 10^{- 3} & = \frac{D \times 640 \times 10^{- 9}}{d} \frac{D}{d} & = \frac{7.2 \times 10^{- 3}}{640 \times 10^{- 9}} = \frac{7.2}{640 \times 10^{- 6}} \end{aligned}$

Now with another monochromatic light $λ^{'}$ , the fringe width $β^{'} = 8.1 mm$ . Hence

$8.1 \times 10^{- 3} = \frac{D \times λ^{'}}{d}$

Putting the value of $\frac{D}{d}$

$\begin{aligned} 8.1 \times 10^{- 3} & = \frac{7.2}{640 \times 10^{- 6}} \times λ^{'} λ^{'} & = \frac{640 \times 8.1 \times 10^{- 9}}{7.2} & = 720 nm \end{aligned}$

Condition for minimum value of the order $(n)$ of bright fringe of shorter wavelength which coincides with that of the longer wavelength

$\begin{aligned} \frac{n_{1} D}{d} λ & = \frac{n_{2} D}{d} λ^{'} \frac{n_{1}}{n_{2}} & = \frac{λ^{'}}{λ} \frac{n_{1}}{n_{2}} & = \frac{720 \times 10^{- 9}}{640 \times 10^{- 9}} \frac{n_{1}}{n_{2}} & = \frac{9}{8} \end{aligned}$

Answering Tips

All quantities must be brought to SI system before substituting them in formula.

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wave-optics Question 16

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Wave Optics

wave-optics Question 16

Answering Tips

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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