Question

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width W2), and a wavelength of λ2 = 496 nm is used. The width of the central bright fringe on the screen is observed to be unchanged. Find W2.

Answer 1

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$, then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Answer 2

The width of the central bright fringe on the screen is observed to be unchanged is $4.48 \times 10^{-6}m$

Calculation of the width:

It is important to apply the concepts i.e. related to interference from two sources. Destructive interference generated the dark fringes.  Dark fringes in the diffraction pattern of a single slit should be found at angles θ i.e. shown below.

$wsim{\theta} = m \lambda$

here

w = width

wavelength

And, m is an integer, m = 1, 2, 3...

So,

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}\\\\w_2 = \frac{4.4\times 10^{-6}496}{487}$

$4.48 \times 10^{-6}m$

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