Question

In a young's double-slit experiment, the 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 maxima to the point where the bright fringes due to both the wavelengths coincide is:

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

Answer 1

Answer: (D)

$7.8mm$

Given,

$d=0.5mm=0.5\times {10}^{-3}m$

$D=150cm=1.5m$

${\lambda }_1=650nm$

${\lambda }_2=520nm$

For common maxima,

$n_1{\lambda }_1=n_2{\lambda }_2$

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}$

$\frac{n_1}{n_2}=\frac{520}{650}=\frac{4}{5}$

Now, $n_1=4,{\lambda }_1=650nm$

Condition of maxima,

$\frac{yd}{D}=n_1{\lambda }_1$

$y=\frac{n_1{\lambda }_{1}D}{d}$

$y=\frac{4\times 650\times {10}^{-9}\times 1.5}{0.5\times {10}^{-3}}=7.8mm$

The correct option is D.