Question

In an experiment the two slits are $0.5$mm apart and the fringes are observed to$100$cm from the plane of the slits. The distance of the $11th$ bright fringe from the 1st bright fringe is $9.72mm$. Calculate the wavelength-

• A. $4.86\times {10}^{-5}cm$
• B. $4.86\times {10}^{-8}cm$
• C. $4.86\times {10}^{-6}cm$
• D. $4.86\times {10}^{-7}cm$

Answer 1

The correct option is A $4.86\times {10}^{-5}cm$

We have,

$d=0.5$ mm

$D=100cm=1000mm$

we know that,

path difference $\left(△x\right)$

$△x=\frac{xd}{D}$

for maxima (bright triangles)

$△x=n\lambda$

where $\lambda \to$ wavelength

$\Rightarrow \frac{xd}{D}=n\lambda$

$\Rightarrow x=\frac{n\lambda D}{d}$

now,

distance $\left(x_1\right)$ & I bright fringe

putting $n=1$

$x_1=\frac{\lambda D}{d}$

distance$\left(x_{11}\right)$ of ${11}^{th}$ bright fringe (putting $n=11$)

$x_{11}=\frac{11\lambda D}{d}$

according to question

$x_{11}-x_1=9.72mm$

$\Rightarrow \frac{11\lambda D}{d}-\frac{\lambda D}{d}=9.72$

$\Rightarrow \frac{10\lambda 0}{d}=9.72$

$\Rightarrow \lambda =\frac{9.72\times d}{10D}$

$=\frac{9.72\times 0.5}{10\times 1000}$

$\Rightarrow \lambda =4.86\times {10}^{-4}mm$

$\lambda =4.86\times {10}^{-5}cm$