Respuesta :
Answer:
[tex]\Delta \lambda=14.3\ nm[/tex]
Explanation:
It is given that,
The number of lines per unit length, N = 900 slits per cm
Distance between the formed pattern and the grating, l = 2.3 m
n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, [tex]\Delta Y=2.98\ mm = 0.00298\ m[/tex]
Let d is the slit width of the grating,
[tex]d=\dfrac{1}{N}[/tex]
[tex]d=\dfrac{1}{900\ cm}[/tex]
[tex]d=1.11\times 10^{-5}\ m[/tex]
For the first wavelength, the position of maxima is given by :
[tex]y_1=\dfrac{L\lambda_1}{d}[/tex]
For the other wavelength, the position of maxima is given by :
[tex]y_2=\dfrac{L\lambda_2}{d}[/tex]
So,
[tex]\Delta \lambda=\dfrac{\Delta y d}{l}[/tex]
[tex]\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}[/tex]
[tex]\Delta \lambda=1.43\times 10^{-8}\ m[/tex]
or
[tex]\Delta \lambda=14.3\ nm[/tex]
So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.
