In a single slit diffraction pattern, a light of wavelength 6000… - Solution | JEE Physics

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In a single slit diffraction pattern, a light of wavelength $6000 \, \text{Å}$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 \, \text{mm}$ when the screen is placed $50 \, \text{cm}$ away from the slits. The width of the slit is $\times 10^{-4} \, \text{m}$ :

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: wavelength $\lambda = 6000 \times 10^{-10} \, \text{m}$ , screen distance $D = 0.5 \, \text{m}$ , and distance between first and third minima $y_3 - y_1 = 0.003 \, \text{m}$ .

Find: slit width $a$ .

For single slit diffraction, minima are given by

a \sin \theta = n\lambda

For small angles,

\sin \theta \approx \tan \theta = \frac{y}{D}

So the position of the $n$ th minimum is

y_n = \frac{n\lambda D}{a}

Detailed Calculation

For the first minimum,

y_1 = \frac{\lambda D}{a}

For the third minimum,

y_3 = \frac{3\lambda D}{a}

Hence the separation between them is

y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}

Given this separation is $0.003 \, \text{m}$ , so

\frac{2\lambda D}{a} = 0.003

Therefore,

a = \frac{2\lambda D}{0.003}

Direct Separation Formula

Since the separation between first and third minima is two minima spacings, one can directly use

y_3 - y_1 = \frac{2\lambda D}{a}

Substituting $\lambda = 6000 \times 10^{-10} \, \text{m}$ and $D = 0.5 \, \text{m}$ ,

a = \frac{2 \cdot 6000 \times 10^{-10} \cdot 0.5}{0.003}

a = 2 \times 10^{-4} \, \text{m}

Therefore, the required numerical value is 2.

Common mistakes

Using the width of the central maximum instead of the separation between the first and third minima is incorrect. Here the given distance is $y_3 - y_1$ , not the central maximum width. Use $y_3 - y_1 = \frac{2\lambda D}{a}$ .
Not converting units consistently leads to error. $6000 \, \text{Å}$ must be written as $6000 \times 10^{-10} \, \text{m}$ and $3 \, \text{mm}$ as $0.003 \, \text{m}$ before substitution.
Taking the third minimum position as the separation directly is wrong. The required distance is between first and third minima, so subtract: $y_3 - y_1$ , not just $y_3$ .

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