Light consisting of two nearly equal wavelengths λ+Δλand...

Light consisting of two nearly equal wavelengths $\lambda+\Delta \lambda$ and $\lambda$ where $\Delta \lambda \ll \lambda$, is incident on a diffraction grating. The slit separation of the grating is $d$ a. Show that the angular separation of these two wavelengths in the $m$th order is $$\Delta \theta=\frac{\Delta \lambda}{\sqrt{(d / m)^{2}-\lambda^{2}}}$$ b. Sodium atoms emit light at $589.0 \mathrm{nm}$ and $589.6 \mathrm{nm} .$ What are the first-order and second-order angular separations (in degrees) of these two wavelengths for a 600 line/mm grating?

Light consisting of two nearly equal wavelengths $\lambda+\Delta \lambda$ and $\lambda$ where $\Delta \lambda \ll \lambda$, is incident on a diffraction grating. The slit separation of the grating is $d$
a. Show that the angular separation of these two wavelengths in the $m$th order is $$\Delta \theta=\frac{\Delta \lambda}{\sqrt{(d / m)^{2}-\lambda^{2}}}$$ b. Sodium atoms emit light at $589.0 \mathrm{nm}$ and $589.6 \mathrm{nm} .$ What are the first-order and second-order angular separations (in degrees) of these two wavelengths for a 600 line/mm grating?

Key Concepts

Angular Dispersion through Differentiation

Angular dispersion refers to the variation in the diffraction angle with respect to changes in wavelength. By differentiating the diffraction grating equation with respect to ?, one obtains an expression that quantifies how a small difference in wavelength (??) leads to a corresponding angular separation (??). This concept is essential for understanding how gratings can resolve closely spaced wavelengths and affects the design and performance of optical instruments that rely on spectral analysis.

Diffraction Grating Equation

This is the fundamental equation describing how light interacts with a grating. It states that constructive interference occurs when the path difference between adjacent slits is an integer multiple of the wavelength, mathematically expressed as d sin(?) = m?, where d is the slit separation, ? is the diffraction angle, m is the order of the spectrum, and ? is the wavelength. This equation is crucial for predicting the positions of the spectral lines produced by the grating.

Transcript

00:01 In the first part of this problem, we have to show that the delta theta equals to delta lambda divided by square root of d divided by m, whole square root square minus lambda square.

00:17 Here this delta theta is the angular expression, delta lambda is the change in the wavelength, the slit expression, m is the order of the fringe, and this lambda is the wayfence of the light.

00:27 Now let's start from the condition of the construction difference in the case of difuritian grating which is a d sine theta equals to m lambda now it can be written as a sine theta equals to m lambda divided by d we call it equation number one now taking the differentiation of this equation we can write d cosine theta d d theta equals to m d lambda.

01:01 Now let's assume that this d theta is approximately equals to delta theta and this d lambda is equal to delta lambda.

01:14 So this can be written as d cosine of theta, delta, theta equals to m delta lambda now this expression can be written as d divided by m cosine of theta equals to theta delta theta equal to delta lambda now we know that this a cosine theta is equal to square root of one minus sine square theta and we call it equation number two where sine theta equals to m lambda divided by d so this can be written as cosine theta equals to square root of 1 minus m lambda divided by d whole square now inserting this value into equation number 2 we can write d divided by m whole square square root into square root of 1 minus m lambda divided by d whole square into delta theta equals to delta lambda now multiplying by this term we can write this this expression as square root of d divided by m whole square minus lambda square into delta theta equals to delta lambda now solving the square in for this delta theta we can write delta theta equals to delta lambda divided by square root of d divided by m whole square minus lambda square hence proved now in part b of this problem we are going to calculate the first order and second order angular expression that is delta theta 1 and delta theta 2 now for delta theta 1 we can write m equals to 1 and so using the equation which we have derived in the above case we can write delta theta 1 equals 2 we have devalued for delta lambda as 500 and it is 0 .6 nanometer minus 589 .9 .0 nanometer divided by square root of vfd value for d as 1 .0 multiplied by 10x power minus 3 divided by 600 divided by 1 whole square minus 500 and 809 .6 multiplied by 10 x power minus 9 whole square meter whole square so this will give the value for this delta theta 1 as delta theta 1 equals to 3 .85 multiply by 10 raise power minus 4 radiance and in degree we can write it as a delta theta 1 equals to 3 .85 multiply by 10 raise power minus 4 radian multiply by 180 degree divided by pyre variance...

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