Question

A diffraction grating with 10^6 lines /m is used to determine the wavelengthof a monochromatic source. the angle of first order diffraction is 30 degree . The wavelength of the source is ? options: a) 1000 nm b) 500 nm c) 400 nm d) 600 nm

Answer 1

Answer:

500 nm

Explanation:

you know that    d sin ( theta )  = n ( wavelenght )

                               sin ( theta ) = wavelenght x ( n / d )

                               sin 30 = wavelenght x 10^6

                               0.5 = wavelenght x 10^6

                          5 x 10^ -3  = wavelenght x 10^6

                   5 x ( 10^ -3 ) x ( 10^ -6 ) = wavelenght

 thus wavelenght  = 500 nm

Answer 2

Answer:

The wavelength of the source will be 500 nm.

Explanation:

We know that:

dsin(θ)  = n(wavelength)

sin (θ) = wavelenght x ($\frac{n}{d}$)

sin 30 = wavelenght x 10⁶

0.5 = wavelenght x 10⁶

5 x 10⁻³  = wavelenght x 10⁶

5 x (10⁻³) x (10⁶) = wavelenght

So, the wavelenght  = 500 nm

Thus, the separation between a location on one wave and the same place on the following wave is known as the wavelength of a wave. It is frequently simplest to measure this from one wave's peak to the next.