Question

In a Young's double slit experiment, the fringe width is found to be $2mm$, when the light of wavelength $6000A$ is used. Find the change in fringe width if the whole apparatus is immersed in water of refractive index $1.33$.

• A. $0.5mm$
• B. $1mm$
• C. $1.5mm$
• D. $2mm$

Answer 1

The correct option is C

$1.5mm$

Step 1: Given

The fringe width is found to be $\beta =$$2mm,$

When the light of wavelength $\lambda =6000A$ is used

The change in fringe width if the whole apparatus is immersed in water of refractive index $\mu =1.33$.

Step 2: Formula used

We know that,

Fringe width, $\beta =\frac{\lambda D}{d}$

$$\begin{gathered} Where\lambda iswavelength \\ Disthedis\tan cebetweentheslitsandscreen \\ disthedis\tan cebetweenslits \end{gathered}$$

When immersed in water, the wavelength becomes, ${\lambda }_1=$ $\frac{\mathrm{\lambda }}{\mathrm{\mu }}$

Where, $$\begin{gathered} \mu istherefractiveindex \\ {\lambda }_{1}isthenewwavelength \end{gathered}$$

${\lambda }_1=$$\frac{\mathrm{\lambda }}{\mathrm{\mu }}$

Where, ${\lambda }_1$ is the wavelength after the whole apparatus is immersed in water.

${\lambda }_1=\frac{6000}{1.33}=4,511.27A$

We know that,

$\beta =\frac{\lambda D}{d}$

$2mm=\frac{\left(6000\right)D}{d}$

$\frac{D}{d}=\frac{2mm}{6000A}$

Step 4: Calculating new fringe width

${\beta }_{new}=\frac{{\lambda }_{1}D}{d}$

Where, ${\beta }_{new}$is the new fringe width.

Substituting the values, we get

${\beta }_{new}=(4511.27A)\left(\frac{2mm}{6000A}\right)$

${\beta }_{new}=1.5mm$

Hence, In Young's double slit experiment, the fringe width is found to be $2mm$, when the light of wavelength $6000A$ is used then the change in fringe width if the whole apparatus is immersed in water of refractive index $1.33$ is. $1.5mm$.

Therefore, option (C) is the correct answer.