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Question: The fringe width in YDSE is \(2.4 \times {10^{ - 4}}m\), when a red light of wavelength \(6400\matho...

The fringe width in YDSE is 2.4×10−4m2.4 \times {10^{ - 4}}m, when a red light of wavelength 6400A∘6400\mathop A\limits^ \circ is used. By how much will it change, if blue of wavelength 4000A∘4000\mathop A\limits^ \circ is used?

Explanation

Solution

In order to this question, first we will rewrite the given value of fringe width and then we will apply the formula of YDSE derivatives, to find the value of Dθ\dfrac{D}{\theta } . And then we will find the change in wavelength.

Formula used:
We will use the formula of Fringe width,
β=xDθ\beta = x\dfrac{D}{\theta }

Complete step by step answer:
Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used.Young’s double-slit experiment helped in understanding the wave theory of light.

Given that: Fringe width, β=2.4×10−4 m\beta = 2.4 \times {10^{ - 4}}\,m
Wavelength of red light is 6400A∘6400\mathop A\limits^ \circ .
As we know that, according to YDSE derivation formula-
β=xDθ\beta = x\dfrac{D}{\theta }
where, β\beta is the fringe width and xx is the wavelength of red light.
2.4 \times {10^{ - 4}}m = 6400 \times {10^{ - 10}}\dfrac{D}{\theta } \\\ \Rightarrow \dfrac{D}{\theta } = \dfrac{{2.4 \times {{10}^{ - 4}}m}}{{6400 \times {{10}^{ - 10}}}} \\\
Now, we will apply the formula of wavelength-
β1=4000×10−10×Dθ ⇒β1=4000×10−10×2.4×10−46400×10−10 ⇒β1=1.5×10−4 {\beta ^1} = 4000 \times {10^{ - 10}} \times \dfrac{D}{\theta } \\\ \Rightarrow {\beta ^1}= 4000 \times {10^{ - 10}} \times \dfrac{{2.4 \times {{10}^{ - 4}}}}{{6400 \times {{10}^{ - 10}}}} \\\ \Rightarrow {\beta ^1}= 1.5 \times {10^{ - 4}} \\\
Now we can calculate the change in wavelength-
β−β1=(2.4−1.5)×10−4 ∴β−β1=0.9×10−4cm \beta - {\beta ^1} = (2.4 - 1.5) \times {10^{ - 4}} \\\ \therefore \beta - {\beta ^1} = 0.9 \times {10^{ - 4}}cm \\\
Hence, the required change in wavelength is 0.9×10−4 cm0.9 \times {10^{ - 4}}\,cm.

Note: The Index of Refraction and Wavelength Light travels in waves, with wavefronts that are perpendicular to the motion direction. The wavefronts are depicted by the green parallel lines in this animation. The motion is shown by the red arrow. When light travels from air to water, it not only slows down but also changes wavelength.