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The width of one of the two slits in a YDSE ....

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x:4 where x is _ _ _ _ .

Solution

Given, $\frac {A_1}{A_2} = \frac {3}{1}$

Now, $\frac {I_1}{I_2} = \frac {A_1 ^2 }{A_2 ^2} = (\frac {3}{1})^2 = \frac {9}{1}$

$\frac {I_{min}}{I_{max}} = \frac {(\sqrt I_1 - \sqrt I_2 )^2 }{(\sqrt I_1 + \sqrt I_2 )^2} = \frac {(3-1)^2}{(3+1)^2}=1:4$

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$