NEET-XII-Physics
17: Light Waves
- #23Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?Ans : Given:
The mirror reflects 64% of the energy or intensity of light.
Let intensity of source = I1.
And intensity of light after reflection from the mirror = I2.
Let a1 and a2 be corresponding amplitudes of intensity I1 and I2.
According to the question,
`` {I}_{2}=\frac{{I}_{1}\times 64}{100}``
`` \Rightarrow \frac{{I}_{2}}{{I}_{1}}=\frac{64}{100}=\frac{16}{25}``
`` \,\mathrm{\,And\,}\frac{{I}_{2}}{{I}_{1}}=\frac{{{a}_{2}}^{2}}{{{a}_{1}}^{2}}``
`` \Rightarrow \frac{{a}_{2}}{{a}_{1}}=\frac{4}{5}``
`` \,\mathrm{\,We\,}\,\mathrm{\,know\,}\text{that}\frac{{I}_{\,\mathrm{\,max\,}}}{{I}_{\,\mathrm{\,min\,}}}=\frac{{\left({a}_{1}+{a}_{2}\right)}^{2}}{{\left({a}_{1}-{a}_{2}\right)}^{2}}``
`` =\frac{{\left(5+4\right)}^{2}}{{\left(5-4\right)}^{2}}``
`` {I}_{\,\mathrm{\,max\,}}:{I}_{\,\mathrm{\,min\,}}=81:1``
Hence, the required ratio is 81 : 1.
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