MCQ
In Young's double slit interference experiment , the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is
- A$3 : 2$
- B$1 : 25$
- ✓$9: 4$
- D$1 : 5$
✓
Answer
Correct option: C.
$9: 4$
c
We know that,
We know that,
$\frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{{{{\left( {\sqrt {\frac{{{\omega _1}}}{{{\omega _2}}} + 1} } \right)}^2}}}{{{{\left( {\sqrt {\frac{{{\omega _1}}}{{{\omega _2}}} - 1} } \right)}^2}}}$
$I_{\max }$ and $I_{\min }$ are maximum and minium intensity
$\omega_{1}$ and $\omega_{2}$ are widths of two slits
$\therefore \,\,\frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{{{{\left( {\sqrt {\frac{1}{{25}}} + 1} \right)}^2}}}{{{{\left( {\sqrt {\frac{1}{{25}}} - 1} \right)}^2}}}$ $\left( {\frac{{{\omega _1}}}{{{\omega _2}}} = \frac{1}{{25}}\operatorname{given} } \right)$
On solving we get,
$\frac{{{I_{\max }}}}{{{I_{\min }}}} = \frac{{\frac{{36}}{{25}}}}{{\frac{{16}}{{25}}}} = \frac{9}{4} = 9:4$
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