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  • (a) Briefly describe the Young’s double slit experiment of interference of light. Drive the expression for fringe width in the pattern. (b) Monochromatic light of wavelength 588 nm is incident from air to water interface. Find the wavel

(a) Briefly describe the Young’s double slit experiment of interference of light. Drive the expression for fringe width in the pattern.

(b) Monochromatic light of wavelength 588 nm is incident from air to water interface. Find the wavelength and speed of the refracted light. The refractive index of water is\frac{4}{3} .

 

 

 

 
 
 
 
 

Answers (1)

a) 

Light from a bright source was made to fall on two silts S1 and S2 made on an opaque screen through slit S. When light waves from two illuminated silts are induced on screen, the path travelled by each wave is different. Thus leads to a phase difference in the two light waves. The path difference is different for each point on the screen. As a result, the intensity is different at all points. This results in the formation of bright dark fringes on the screen.

When S1 and S2 are two coherent sources and for an arbitrary point P on the line GG' to correspond to the maximum we must have

S_{2}P-S_{1}P=n\lambda; n = 0,1,2...

Now 

\left ( S_{2}P \right )^{2}-\left ( S_{1}P \right )^{2}=\left [ D^{2}+\left ( x +\frac{d}{2} \right )^{2} \right ]-\left [ D^{2}+\left ( x -\frac{d}{2} \right )^{2} \right ]=2xdwhere S_{1}S_{2}=d and OP = x

S_{2}P-S_{1}P= \frac{2xd}{S_{2}P+S_{1}P}

If x, d <<D, then negligible error will be introduced if S_{2}P+S_{1}P is replaced by 2D.

\therefore S_{2}P-S_{1}P\cong \frac{xd}{D}

hence we will have constructive interference resulting in a bright region when

x = x_{n}= \frac{n\lambda D}{d}; n=0,\pm 1,\pm 2....                         .......(1)

We will have a dark region near

x = x_{n}= \left ( n+\frac{1}{2} \right )\frac{\lambda D}{d}; n=0,\pm 1,\pm 2....                .......(2)

Fringe width \beta = x_{n+1}-x_{n}

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

b) 

\\\mu_w=\frac{C_0}{C_w}=\frac{\nu\lambda_0}{\nu\lambda_w}\\\\\lambda_w=\frac{\lambda_0}{\mu_w}

\frac{588\times10^{-9}}{\frac{4}{3}}=441nm

C_w=\frac{C_0}{\mu_w}=\frac{3\times10^{8}}{4/3}=2.25\times10^{8}m/sec

Posted by

Safeer PP

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