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Question: Two slits are separated by a distance of0.5mm and illuminated with light of wavelength \(\lambda \) ...

Two slits are separated by a distance of0.5mm and illuminated with light of wavelength λ\lambda = 6000 A0{A^0}.If the screen is placed at 2.5m from the slits. The distance of the third bright fringe from the centre will be:
A) 1.5mm
B) 3mm
C) 6mm
D) 9mm

Explanation

Solution

According to the young’s double slit experiment the distance of the third bright fringe from the central fringe will be given as:
y=3Dλdy = \dfrac{{3D\lambda }}{d} (Where D is the distance of the screen from the slits, d is the distance between the two slits, λ\lambda is the wavelength of the monochromatic light used for the experiment, y is the distance of fringe from the screen).
Using the above formula we will calculate the distance of the third bright fringe.

Complete step by step solution:
Let’s define a few terms relation to the double slit experiment:
Interference of light: the phenomenon of non uniform distribution of energy in the medium due to superposition of two light waves is called Interference of light.
Fringe: On the interference of light waves, alternate dark and bright bands of light are called fringes.
Now comes to the calculation part:
We have,
⇒y=3Dλd\Rightarrow y = \dfrac{{3D\lambda }}{d}
On substituting all the parameters of the equation given in the question:
⇒y=3×2.5×6000×10−100.5×10−3\Rightarrow y = \dfrac{{3 \times 2.5 \times 6000 \times {{10}^{ - 10}}}}{{0.5 \times {{10}^{ - 3}}}}
Now we will perform the simple mathematical calculation to obtain y.
⇒y=15×6000×10−1010−3 ⇒y=9×104×10−10×103  \Rightarrow y = \dfrac{{15 \times 6000 \times {{10}^{ - 10}}}}{{{{10}^{ - 3}}}} \\\ \Rightarrow y = 9 \times {10^4} \times {10^{ - 10}} \times {10^3} \\\
When bases are same powers are added, here the same base is 10 so we will add all the powers;
⇒y=9×10−3 ∴y=9mm  \Rightarrow y = 9 \times {10^{ - 3}} \\\ \therefore y = 9mm \\\
Option (D) is correct.

Note: Fringes can be dark or bright. By dark or bright we mean that area where intensity of light is more forms the bright fringe and the area on the screen where the intensity of the light is less shows dark fringe. When white light is used instead of monochromatic light ( light having a single wavelength), different patterns of interference are seen in one experiment of the same screen.