A beam of light consisting of two wavelengths 7000 and 5500 is us… - Solution | JEE Physics

NVAMediumJEE 2023Interference (Young's Experiment)

A beam of light consisting of two wavelengths $7000 \, \text{\AA}$ and $5500 \, \text{\AA}$ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is $2.5 \, \text{mm}$ and the distance between the place of slits and the screen is $150 \, \text{cm}$ . The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is $n \times 10^{-5} \, \text{m}$ . The value of $n$ is

Answer

Correct answer:462

Step-by-step solution

Standard Method

Given: Two wavelengths are $7000 \, \text{\AA}$ and $5500 \, \text{\AA}$ . Slit separation is $d = 2.5 \, \text{mm}$ and screen distance is $D = 150 \, \text{cm}$ .

Find: The least distance from the central fringe where bright fringes for both wavelengths coincide, and hence the value of $n$ .

For coincidence of bright fringes,

n_1 \beta_1 = n_2 \beta_2

Using $\beta = \frac{\lambda D}{d}$ ,

\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14}

Hence, the ** $11$ **th maximum of $7000 \, \text{\AA}$ coincides with the ** $14$ **th maximum of $5500 \, \text{\AA}$ .

The least distance from the central fringe is

y = n_1 \frac{\lambda_1 D}{d}

So,

y = \frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}} = 462 \times 10^{-5} \, \text{m}

Therefore, $n = 462$ .

Common mistakes

Using the ratio of wavelengths in the wrong order. From $n_1 \beta_1 = n_2 \beta_2$ and $\beta \propto \lambda$ , one must write $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$ . Reversing this gives the wrong pair of coinciding maxima.
Forgetting unit conversion for angstrom, centimetre, or millimetre. Convert $7000 \, \text{\AA}$ and $5500 \, \text{\AA}$ to metres, $150 \, \text{cm}$ to metres, and $2.5 \, \text{mm}$ to metres before substitution.
Using the fringe width formula incorrectly as $\beta = \frac{\lambda d}{D}$ . The correct expression in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$ .

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