Question

Answer the following questions:
(A) In a double slit experiment using light of wavelength $600nm$, the angular width of the fringe formed on a distant screen is ${0.1^ \circ }$. Find the spacing between the two slits.
(B) Light of wavelength $5000{A^ \circ }$ propagating in air gets partly reflected from the surface of water. How will the wavelength and frequencies of the reflected and refracted light be affected?

Answer 1

In this question, you have given the angular width of the fringe , use the formula for angular width to find the spacing between two slits. Recall the formula for the frequency and also remember that Wavelength remains the same in the reflected wave, but decreases in the refracted wave.

Formula used:
Angular width of the fringe,
$\theta = \dfrac{\lambda }{d}$
Where, $\lambda$ is the wavelength of light
$d$ is the spacing between the slits
Frequency,
$\nu = \dfrac{c}{\lambda }$
Where, $c$ is the speed of light in air
$\lambda$ is the wavelength of light

Complete step by step solution:
(A) In the question, the wavelength of the light is given as,
$\lambda = 600nm$
Convert it into the SI unit that is, in meters
$\Rightarrow \lambda = 600 \times {10^{ - 9}}m$
And, also given the angular width of the fringe formed,
$\theta = 0.1^\circ$
On converting it into radian, we get
$\Rightarrow \theta = 0.1 \times \dfrac{\pi }{{180}}rad$
Now, using the formula for angular width of the fringe,
$\theta = \dfrac{\lambda }{d}$
By using this formula we can find the spacing between the slits,
$\Rightarrow d = \dfrac{\lambda }{\theta }$
On putting all the values available we get,
$\Rightarrow d = \dfrac{{600 \times {{10}^{ - 9}}}}{{0.1 \times \dfrac{\pi }{{180}}}}$
$\Rightarrow d = \dfrac{{600 \times {{10}^{ - 9}} \times 180 \times 10}}{{3.14}}$
On further solving, we get the spacing between the slits,
$\Rightarrow d = 0.343 \times {10^{ - 3}}m$

Hence, the spacing between the two slits is $343\mu m$.

(B) In the question, the wavelength of the light is given as,
$\lambda = 5000{A^ \circ }$
Convert it into the SI unit that is, in meters
$\Rightarrow \lambda = 5000 \times {10^{ - 10}}m$
The frequency of reflected and refracted light is the same.
Therefore,
The formula for frequency is given by,
$\nu = \dfrac{c}{\lambda }$
On putting all the values available,
$\Rightarrow \nu = \dfrac{{3 \times {{10}^8}}}{{5 \times {{10}^{ - 7}}}}$
On further calculating, we get the frequency as,
$\Rightarrow \nu = 6 \times {10^{14}}Hz$
As frequency of reflected and refracted light is the same, this is the required frequency.
We know that the formula for the refractive index of water,
$\mu = \dfrac{c}{v}$
As we know the refractive index for water is given by $\dfrac{4}{3}$
On putting this value in above equation,
$\Rightarrow \dfrac{4}{3} = \dfrac{{3 \times {{10}^8}}}{v}$
On further solving, we get
$\Rightarrow v = 2.25 \times {10^8}m/s$
Therefore, the wavelength of the refracted light,
$\lambda = \dfrac{{2.25 \times {{10}^8}}}{{6 \times {{10}^{14}}}}$
On solving, we get
$\Rightarrow \lambda = 0.375 \times {10^{ - 6}}m$

Therefore, the wavelength of the refracted light is $375nm$.

Note: The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles; moreover, it displays the fundamentally probabilistic nature of quantum mechanical phenomena.