Question 86: Wave Optics

Question

A monochromatic light of wavelength $6000 ~\mathring{A}$ is incident on the single slit of width $0.01 \mathrm{~mm}$. If the diffraction pattern is formed at the focus of the convex lens of focal length $20 \mathrm{~cm}$, the linear width of the central maximum is :

Accepted Answer

To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum. The angle, $ 	heta $, at which the first minimum occurs is given by:
$$ a \sin(	heta) = m\lambda $$
Where:
$ a $ is the width of the slit
$ \lambda $ is the wavelength of monochromatic light
$ m $ is the order number of the minimum (for the first minimum, $ m = \pm 1 $)
In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen. So we will use $ m = 1 $ and $ m = -1 $ which correspond to the first minima on either side of the central peak.
Given the width of the slit $ a = 0.01 $ mm, which we need to convert to meters for consistency with the wavelength:
$$ a = 0.01 	imes 10^{-3} 	ext{m} $$
And the wavelength $ \lambda = 6000 \mathring{A} $, also converting to meters:
$$ \lambda = 6000 	imes 10^{-10} 	ext{m} $$
We're using a convex lens of focal length $ f = 20 $ cm to project the pattern onto a screen. We'll need to convert the focal length to meters as well:
$$ f = 20 	imes 10^{-2} 	ext{m} $$
We can calculate the angle $ 	heta $ needed for the first minima using the approximation of small angles, where $ \sin(	heta) \approx 	heta $:
$$ a 	heta = m\lambda $$
Now, solve for $ 	heta $ for the first minimum (m = 1):
$$ 	heta = \frac{\lambda}{a} $$
Plugging in the values:
$$ 	heta = \frac{6000 	imes 10^{-10}}{0.01 	imes 10^{-3}} $$
$$ 	heta = \frac{6000 	imes 10^{-10}}{10^{-5}} $$
$$ 	heta = 6000 	imes 10^{-5} $$
Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $ f $ and the angle $ 	heta $:
$$ y = 2f	heta $$
Plugging the focal length and calculated theta:
$$ y = 2 	imes 20 	imes 10^{-2} 	imes 6000 	imes 10^{-5} $$
$$ y = 40 	imes 10^{-2} 	imes 6000 	imes 10^{-5} $$
$$ y = 240 	imes 10^{-3} 	ext{m} $$
$$ y = 24 	imes 10^{-2} 	ext{m} $$
$$ y = 24 	ext{mm} $$
This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.

Answer

$12 \mathrm{~mm}$

Answer

$24 \mathrm{~mm}$

Answer

$60 \mathrm{~mm}$

Answer

$120 \mathrm{~mm}$

Wave Optics

A monochromatic light of wavelength 6000 ~\mathring{A} is incident on the single slit of width 0.01 \mathrm{~mm}. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 \mathrm{~cm}, the linear width of the central maximum is :

Choose the correct answer: