Subjective Type

A light with wavelength $$\lambda = 0.60 \mu m$$ falls normally on a diffraction grating inscribed on a plane surface of a plano-convex cylindrical glass lens with curvature radius $$R = 20 cm$$. the period of grating is equal to $$d = 6.0 \mu m$$. Find the distance between the principal maxima of first order located symmetrically in the focal plane of that lens.

Solution

For the lens $$\dfrac{1}{f} = (n - 1) \left(\dfrac{1}{R} - \dfrac{1}{\infty} \right) $$ or $$f = \dfrac{R}{n - 1}$$

For the grating $$d \sin \theta_1 = \lambda $$ or $$\sin \theta_1 = \dfrac{\lambda}{d}$$

$$cosec \theta_1 = \dfrac{d}{\lambda}, \cot \theta_1 = \sqrt{\left(\dfrac{d}{\lambda}\right)^2 - 1}$$

$$\tan \theta_1 = \dfrac{1}{\sqrt{\left(\dfrac{d}{\lambda}\right)^2 - 1}}$$

Hence the distance between the two symmetrically placed first order maxima $$= 2 \ f \tan \theta_1 = \dfrac{2R}{(n - 1) \sqrt{\left(\dfrac{d}{\lambda}\right)^2 - 1]}}$$

On putting $$R = 20, n = 1.5 , d = 6.0 \mu m$$
$$\lambda = 0.60 \mu m$$ we get $$8.04 \ cm$$.

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