
A ray of light of intensity I is incident on a parallel glass slab at point $A$ as shown in diagram. It…

- 49: 1
- 7: 1
- 4: 1
- 8: 1
Solution

From figure $\mathrm{I}_{1}=\frac{\mathrm{I}}{4}$ and $\mathrm{I}_{2}=\frac{9 \mathrm{I}}{64}$ $\begin{aligned} \text { By using } \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }} &=\left(\frac{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}+1}{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}-1}\right)^{2} \\ &=\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)^{2}=\frac{49}{1} \end{aligned}$
Asked in: TEST SERIES MHT-CET Full Test 6
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