Question 182: Wave Optics

Question

In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $ 2 : 1 $ and $1: 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ( $D_1 / D_2$ ) is $\_\_\_\_$。

Accepted Answer

In two Young's Double Slit Experiments (YDSE) the fringe widths are equal ( $\beta_1=\beta_2$ ).
Ratio of slit separations (d) is $\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{2}{1}=2$
Ratio of wavelengths ( $\lambda$ ) is $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$
The formula for fringe width ( $\beta$ ) in a YDSE setup is:
$$ \beta=\frac{\lambda D}{d} $$
where:
$\lambda$ is the wavelength of light.
D is the distance between the slits and the screen.
d is the separation between the slits.
$$ \beta_1=\beta_2 \Rightarrow \frac{\lambda_1 \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\lambda_2 \mathrm{D}_2}{\mathrm{~d}_2} $$
$$ \Rightarrow \frac{\mathrm{D}_1}{\mathrm{D}_2}=\left(\frac{\lambda_2}{\lambda_1}ight) 	imes\left(\frac{\mathrm{d}_1}{\mathrm{~d}_2}ight) $$
Since $\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$, so, $\frac{\lambda_2}{\lambda_1}=2$
Putting the values into the equation :
$$ \frac{\mathrm{D}_1}{\mathrm{D}_2}=(2) 	imes(2)=4 $$
The corresponding ratio of the distances between the slits and the respective screens is $4: 1$.
Hence, the correct answer is $\mathbf{4}$.

Wave Optics

Integer / Numeric Answer Type