Optics Ques 140
- In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern
(2000, 2M)
(a) the intensities of both the maxima and the minima increases
(b) the intensity of the maxima increases and the minima has zero intensity
(c) the intensity of maxima decreases and that of minima increases
(d) the intensity of maxima decreases and the minima has zero intensity
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Answer:
Correct Answer: 140.(a)
Solution:
- In interference we know that
$ \begin{array}{ll} & I _{\text {max }}=\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2} \\ \text { and } \quad & I _{\text {min }}=\left(\sqrt{I _1} \sim \sqrt{I _2}\right)^{2} \end{array} $
Under normal conditions (when the widths of both the slits are equal)
$I _1 \approx I _2=I \text { (say) }$
$\therefore \quad I _{\max } =4 I \text { and } I _{\min }=0$
When the width of one of the slits is increased. Intensity due to that slit would increase, while that of the other will remain same. So, let :
$I _1=I$
$ \begin{array}{ll} \text { and } & I _2=\eta I \\ \text { Then, } & I _{\max }=I(1+\sqrt{\eta})^{2}>4 I \\ \text { and } & I _{\min }=I(\sqrt{\eta}-1)^{2}>0 \end{array} $
$\therefore$ Intensity of both maxima and minima is increased.
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