Question

In Young’s double-slit experiment, slits are separated by $0.5mm$, and the screen is placed $150cm$ away. A beam of light consisting of two wavelengths, $650nm$ and$520nm$, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

• A. $1.56mm$
• B. $7.8mm$
• C. $9.75mm$
• D. $15.6mm$

Answer 1

The correct option is B

$7.8mm$

Step 1: Given information

Separation between slits$\left(D\right)$$=$ $0.5mm$

The distance between the screen$\left(d\right)$$=$ $150cm$

A beam of light consisting of two wavelengths$\left({\lambda }_{1}and{\lambda }_2\right)$ are $650nm$ and$520nm$, respectively.

Step 2: Formula

Fringe width formula

$$\begin{gathered} \mathrm{\beta }=\frac{\mathrm{\lambda D}}{\mathrm{d}}\mathrm{where}\mathrm{\lambda }\mathrm{is}\mathrm{wavelength},\mathrm{D}\mathrm{is}\mathrm{seperation}\mathrm{between}\mathrm{slit} \\ \mathrm{d}\mathrm{is}\mathrm{distance}\mathrm{between}\mathrm{slit}\mathrm{and}\mathrm{screen}. \end{gathered}$$

Step 3: Calculate the shortest distance between the shared center maximum and the place where both wavelengths' bright fringes intersect

For${\lambda }_1,y=\frac{m{\lambda }_{1}D}{d}$

For ${\lambda }_2,y=\frac{n{\lambda }_{2}D}{d}$

$y=\frac{m{\lambda }_{1}D}{d}=\frac{n{\lambda }_{2}D}{d}$

$\frac{m}{n}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{4}{5}$, Where $m$ and $n$ are positions of fringes.

Step 3: Solving for ${\lambda }_1$:

For${\lambda }_1$,

$$\begin{gathered} y=\frac{m{\lambda }_{1}D}{d} \\ \Rightarrow y=4\frac{\left(150\times {10}^2\right)\times \left(650\times {10}^{-9}\right)}{\left(0.5\times {10}^{-3}\right)} \end{gathered}$$

Therefore, the required distance is $7.8nm$.

Hence, the correct option is (B).