In the arrangement shown in figure, wavelength of light used is . The distance between slits S_1 and S_2 is . The distance between S_3 and S_4 is . If the ratio of maximum to minimum intensity received on screen P is K. Then 10 K is

JEE MainPhysicsModerateWave Optics

In the arrangement shown in figure, wavelength of light used is
$\lambda$ . The distance between slits S_1 and S_2 is $d(d < < D)$ . The
distance between S_3 and S_4 is $u = \frac{\lambda D}{3d}$ . If the
ratio of maximum to minimum intensity received on screen P is K. Then 10
K is

Show answer and solution

If intensity of light incident on slits S_1 and S_2 is

I_{0}

The intensity through S_4 will be

I_{1} = 4I_{0}

The intensity through S_3 will be

I_{2} = 4I_{0}\cos^{2}(\frac{\phi}{2}) = I_{0}

K = \frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{4I_{0}} + \sqrt{I_{0}})}^{2}}{{(\sqrt{4I_{0}} - \sqrt{I_{0}})}^{2}} = {(\frac{3}{1})}^{2} = \frac{9}{1}

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx

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