NUMERICAL 1 [2082]

Solution

Given,

No. of lines per unit length (N) = 500 lines / mm = 500 x 10³ lines/m

Wavelength [latex](\lambda)[/latex] =?

Order of diffraction (n) = 2

Angle [latex](\theta)[/latex] = 30^o

We know that,

[latex]d = \frac{1}{N}[/latex]

Or,  [latex]d = \frac{1}{500\times 10^3} = 2\times 10^{-6}m[/latex]

Now, dSin [latex]\theta = n\lambda[/latex]

Or, dSin[latex]30^o = 2\lambda[/latex]

Or, [latex]\lambda = \frac{d sin30^o}{2}[/latex]

Or, [latex]\lambda = \frac{(2\times 10^{-6})\times 0.5}{2}[/latex]

Or, [latex]\lambda = 5\times 10^{-7}[/latex]  m

NUMERICAL 2 [2081 GIE ‘B’]

Solution

Given,

No. of lines per unit length (N) = 600 lines / mm = [latex]600\times 10^3\ lines/m[/latex]

Wavelength [latex](\lambda) = 700 nm = 700\times 10^{-9}\ m[/latex]

Order of diffraction (n) = 1

Angle [latex](\theta) =?[/latex]

We know that,

[latex]d = \frac{1}{N}[/latex]

Or, [latex]d = \frac{1}{600\times 10^3} = 1.667\times 10^{-6}\ m[/latex]

Now, dSin[latex]\theta  = n\lambda[/latex]

Or, [latex]sin\theta = \frac{n\lambda}{d}[/latex]

Or, [latex]sin\theta = \frac{1\times (700\times 10^{-9})}{(1.667\times 10^{-6})}[/latex]

Or, [latex]sin\theta = 0.42[/latex]

Or, [latex]\theta = sin^{-1}(0.42) = 24.83^o[/latex]

Next Part

Maximum number of orders (n) =?

We know, for maximum diffraction, [latex]sin\theta = 1[/latex]

So, [latex]d sin\theta = n\lambda[/latex]

Or, [latex](1.667\times 10^{-6})\times 1 = n\times (700\times 10^{-9})[/latex]

Or, [latex]n = \frac{1.667\times 10^{-6}}{700\times 10^{-9}}[/latex]

            = 2.38 [latex]\approx 2[/latex]

NUMERICAL 3 [2081 B/C]