Question

Suppose that Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes near the center of the interference pattern?

Short answer

The bright fringes near the center of the interference pattern are apart of 2.25 mm.

Step-by-step solution

Write the given data from the question

Blue-green light Wavelength, $\lambda =500\text{nm}$.

The slit separation, $d=1.20\text{mm}$.

The screen distance from the slit, $D=5.4\text{m}$.

Determine the formulas to calculate how far apart the bright fringes near the center of the interference pattern

The condition for the maxima in Young’s experiment is given as follows.

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …… (1)

Here, d is the distance between the slits, $\mathit{\lambda }$is the wavelength, mis the order, and$\mathit{\theta }$is the angular separation.

Calculate how far apart the bright fringes near the center of the interference pattern

The maximum vertical distance from the center of the pattern is given by,

$\tan \theta \approx \sin \theta =\frac{y_m}{D}$

Substitute $\frac{y_m}{D}$for $\sin \theta$into equation (1).

$$\begin{gathered} d\left(\frac{y_m}{D}\right)=m\lambda \\ {y}_m=\frac{m\lambda D}{d} \end{gathered}$$

Substitute 1 for m, 54 m for D, 500 nm for $\lambda$and $1.20\text{mm}$for d into above equation.

$$\begin{gathered} y_m=\frac{1\times 500\times {10}^{-9}\times 5.4}{1.2\times {10}^{-3}} \\ =\frac{27\times {10}^{-7}}{1.2\times {10}^{-3}} \\ =22.5\times {10}^{-4}\text{m} \\ =2.25\text{mm} \end{gathered}$$.

Hence the bright fringes near the center of interference pattern are apart of 2.25 mm.