Dealing with Reflections and Refractions.

A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25% and transmits the remaining. Find the ratio of maximum and minimum intensities (i.e., I max I min \frac{I_{\max}}{I_{\min}} ) in the interference pattern formed by the two beams obtained after one reflection at each plate.


The answer is 49.

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1 solution

Each plate reflects 25% and transmits(refracts) 75% .

Intensity of incident ray (Initial intensity I ) after each reflection and refraction is shown in figure.

Interference pattern will take place in between rays 1 and 2 .

I 1 = I 4 a n d I 2 = 9 I 64 \Rightarrow \ \ \ \ I_{1}=\frac{I}{4} \ and \ I_{2}=\frac{9I}{64}

I m a x I m i n = ( I 1 + I 2 I 1 I 2 ) 2 = 49 1 = 49 \therefore \quad \quad \quad \quad \frac { { I }_{ max } }{ { I }_{ min } } ={ \left( \frac { \sqrt { { I }_{ 1 } } +\sqrt { { I }_{ 2 } } }{ \sqrt { { I }_{ 1 } } -\sqrt { { I }_{ 2 } } } \right) }^{ 2 }=\frac { 49 }{ 1 } =\quad \boxed{49}

Bhai language clear likha karo

Aishwary Omkar - 6 years, 1 month ago

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which part is not clear ??

Harshvardhan Mehta - 6 years, 1 month ago

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.............................. after one reflection at each plate?????????

Aishwary Omkar - 6 years, 1 month ago

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