As shown in the figure, in Young's double slit experiment, a thin plate of thickness t = 10 m and refractive index = 1.2 is inserted in front of slit S1. The experiment is conducted in air ( = 1) and…

Given: A thin plate of thickness t = 10 10^-6 m, refractive index = 1.2, and light of wavelength = 500 10^-9 m are used. Find: The value of x in the shift x 0. When a thin plate is introduced in front of one slit in Young's double slit experiment, the fringe shift is x = t( - 1) 0 Substituting the given values, x = 10 10^-6(1.2 - 1)5 10^-7 0 x = 10 10^-6 0.25 10^-7 0 x = 2 10^-65 10^-7 0 = 4 0 Hence, x = 4. Therefore, the value of x is 4. Given: The shift is written as x 0. Find: The numerical value of x. Using the fringe shift relation directly, x = t( - 1) Now, x = 10 10^-6 0.2500 10^-9 x = 2 10^-65 10^-7 = 4 Therefore, the value of x is 4.

As shown in the figure, in Young's double slit experiment, a thin… - Solution | JEE Physics

NVAEasyJEE 2023Interference (Young's Experiment)

As shown in the figure, in Young's double slit experiment, a thin plate of thickness $t = 10 \, \mu\text{m}$ and refractive index $\mu = 1.2$ is inserted in front of slit $S_1$ . The experiment is conducted in air ( $\mu = 1$ ) and uses a monochromatic light of wavelength $\lambda = 500 \, \text{nm}$ . Due to the insertion of the plate, central maxima is shifted by a distance of $x\beta_0$ . $\beta_0$ is the fringe-width before the insertion of the plate. The value of the $x$ is _____.

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: A thin plate of thickness $t = 10 \times 10^{-6} \, \text{m}$ , refractive index $\mu = 1.2$ , and light of wavelength $\lambda = 500 \times 10^{-9} \, \text{m}$ are used.

Find: The value of $x$ in the shift $x\beta_0$ .

When a thin plate is introduced in front of one slit in Young's double slit experiment, the fringe shift is

\Delta x = \frac{t(\mu - 1)}{\lambda}\beta_0

Substituting the given values,

\Delta x = \frac{10 \times 10^{-6}(1.2 - 1)}{5 \times 10^{-7}}\beta_0

\Delta x = \frac{10 \times 10^{-6} \times 0.2}{5 \times 10^{-7}}\beta_0

\Delta x = \frac{2 \times 10^{-6}}{5 \times 10^{-7}}\beta_0 = 4\beta_0

Hence, $x = 4$ .

Therefore, the value of $x$ is $4$ .

Direct Ratio Method

Given: The shift is written as $x\beta_0$ .

Find: The numerical value of $x$ .

Using the fringe shift relation directly,

x = \frac{t(\mu - 1)}{\lambda}

Now,

x = \frac{10 \times 10^{-6} \times 0.2}{500 \times 10^{-9}}

x = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} = 4

Therefore, the value of $x$ is $4$ .

Common mistakes

Using $\mu$ instead of $(\mu - 1)$ in the shift formula is incorrect because the extra optical path depends on the excess refractive index over air. Use $\Delta x = \frac{t(\mu - 1)}{\lambda}\beta_0$ .
Not converting $\mu\text{m}$ and $\text{nm}$ into SI units can lead to a wrong ratio. Convert $10 \, \mu\text{m} = 10 \times 10^{-6} \, \text{m}$ and $500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}$ before substitution.
Confusing the fringe shift with fringe width is incorrect. Here the shift is $x\beta_0$ , so after finding $\Delta x = 4\beta_0$ , compare coefficients to get $x = 4$ .

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