Question 104: Waves Optics

Question

The width of fringe is $$2 \mathrm{~mm}$$ on the screen in a double slits experiment for the light of wavelength of $$400 \mathrm{~nm}$$. The width of the fringe for the light of wavelength 600 $$\mathrm{nm}$$ will be:

Accepted Answer

In the double-slit experiment, the fringe width ($\beta$) is given by the formula :
$
\beta = \frac{\lambda D}{d}
$
where:
$\lambda$ is the wavelength of the light,
$D$ is the distance between the screen and the double-slit,
$d$ is the separation between the two slits.
If we are considering a change in wavelength but the fringe width is changing and the setup of the experiment (the values of $D$ and $d$) stays the same, we can see that the fringe width is directly proportional to the wavelength. This is because $D$ and $d$ are constants in this case, so we can write :
$
\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}
$
Here, the given wavelengths are $\lambda_1 = 400 \, 	ext{nm}$ and $\lambda_2 = 600 \, 	ext{nm}$, and the given fringe width for the light of wavelength $400 \, 	ext{nm}$ is $\beta_1 = 2 \, 	ext{mm}$. We are asked to find the fringe width $\beta_2$ for the light of wavelength $600 \, 	ext{nm}$.
Substituting the given values into the proportionality equation, we get :
$
\frac{2 \, 	ext{mm}}{\beta_2} = \frac{400 \, 	ext{nm}}{600 \, 	ext{nm}}
$
Solving this equation for $\beta_2$ gives:
$
\beta_2 = 2 \, 	ext{mm} 	imes \frac{600 \, 	ext{nm}}{400 \, 	ext{nm}} = 3 \, 	ext{mm}
$

Answer

4 mm

Answer

1.33 mm

Answer

2 mm

Answer

3 mm

Waves Optics

The width of fringe is $2 \mathrm{~mm} on the screen in a double slits experiment for the light of wavelength of 400 \mathrm{~nm}. The width of the fringe for the light of wavelength 600 \mathrm{nm}$ will be:

Choose the correct answer: