TS EAMCET · Physics · Wave Optics
A parallel beam of light of intensity \(I_0\) is incident on a coated glass plate. If \(25 \%\) of the incident light is reflected from the upper surface and \(50 \%\) of light is reflected from the lower surface of the glass plate, the ratio of maximum to minimum intensity in the interference region of the reflected light is
- A \(\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2\)
- B \(\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2\)
- C \(\frac{5}{8}\)
- D \(\frac{8}{5}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2\)
Step-by-step Solution
Detailed explanation
According to question, we can draw the following diagram \( I_1=\frac{I_0}{4} \Rightarrow I_2=\frac{3}{8} I_0 \) We know that,…
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