Question

A diffraction grating is used at normal incidence gives a line 5400ao in a certain order superposed on the violet line (4050ao) of the next higher order. If the angle of diffraction is 30o, how many lines per cm are there in grating?.

Answer 1

The number of lines  per cm in the grating is 30864 lines per cm

Given that;

λ1 = 5400A° and λ2 = 4050A°

The angle of diffraction is 30 degree

To find;

Number of lines per cm are there in grating

Solution;

Since both, the wavelengths are superimposed on each other then,

Angle of diffraction of both the wavelengths will be equal to each other.

θ1  = θ2

sinθ1 = sinθ2 = sin30° = 0.5

Now, for 1st wavelength λ1  = (a+b)sinθ1 = nλ1....(1)

for 2nd wavelength = (a+b)sinθ2 = (n+1)λ2....(2)

dividing (2) by (1) we get,

1 = $\frac{n+1}{n}$x λ1 /  λ2

putting the values of λ1 and  λ2 we get,

1 = $\frac{n+1}{n}$ x $\frac{4.05 X10^{-7} }{5.40X 10^{-7} }$

n = n+1 x $\frac{3}{4}$

4n = 3n + 3

n = 3

Therefore, N = $\frac{1}{a + b}$ = sinθ1/nλ1

N = $\frac{0.5}{3 X 5.40 X 10^{-7} }$ = 30864 lines per cm

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Answer 2

Answer:

There are about 1543 lines per cm in the grating.

Explanation:

To find the number of lines per cm in the grating, we need to use the formula:

          d sin theta = m lambda

where d is the distance between the lines in the grating, theta is the angle of diffraction, m is the order of the spectrum, and lambda is the wavelength of the light.

We are given the angle of diffraction (30 degrees) and the wavelengths of the two superposed lines (5400 angstroms and 4050 angstroms), but we do not know the order of the spectrum for either line. We need to find the values of m for both lines, and then use the formula to solve for d.

One possible way to find the values of m is to use the fact that the two lines are superposed, which means that they have the same angle of diffraction. This implies that:

                    m1 lambda1 = m2 lambda2

where m1 and m2 are the orders of the spectrum for the two lines, and lambda1 and lambda2 are their wavelengths. We can use this equation to find the ratio of m1 and m2, and then use trial and error to find the possible values of m1 and m2 that satisfy the equation.

For example, if we assume that m1 = 1, then we can find m2 by:

m2 = m1 lambda1 / lambda2

m2 = 1 *$\frac{ 5400}{4050}$

m2 = 1.333

Since m2 has to be an integer, this value is not possible. We can try other values of m1 until we find a pair of integers that work. One possible pair is m1 = 3 and m2 = 4, which gives:

m1 lambda1 = m2 lambda2

3 * 5400 = 4 * 4050

16200 = 16200

This pair satisfies the equation, so we can use them to find d by:

d = m1 lambda1 / sin theta

d = $\frac{3 * 5400}{sin 30}$

d = $\frac{32400}{0.5}$

d = 64800 angstroms

To convert this to lines per cm, we need to divide by 10^8 and take the reciprocal:

lines per cm = $\frac{10^{8} }{d}$

lines per cm = $\frac{10^{8} }{64800}$

lines per cm = 1543.21

Therefore, there are about 1543 lines per cm in the grating.

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