A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. It undergoes partial reflection and refraction. At each reflection 25 % of incident energy is reflected. The rays A B and A' B' undergo interference. The ratio I max / I min is

Question

A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. It undergoes partial reflection and refraction. At each reflection 25 % of incident energy is reflected. The rays A B and A' B' undergo interference. The ratio I max / I min is (A) 4: 1 (B) 8: 1 (C) 7: 1 (D) 49: 1

Tardigrade Admin · Accepted Answer

From figure I1=(I/4) and I2=(9 I/64) ⇒ (I2/I1)=(9/16) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/46b60cae86eb7e2e748470e5fbdc1688-.png" /> By using (I max /I min )=((√(I2/I1)+1/√(I2)I1-1)) =((√(9/16)+1/√(9)16-1))=(49/1)

4 : 1

8 : 1

7 : 1

49 : 1

From figure $I_{1} = \frac{I}{4}$ and $I_{2} = \frac{9 I}{64} \Rightarrow \frac{I _{2}}{I _{1}} = \frac{9}{16}$

By using $\frac{I _{m a x}}{I _{m i n}} = (\frac{\frac{I _{2}}{I _{1}} + 1}{\frac{I _{2}}{I _{1}} - 1})$
$= (\frac{\frac{9}{16} + 1}{\frac{9}{16} - 1}) = \frac{49}{1}$

Q. A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A′B′ undergo interference. The ratio Imax​/Imin​ is

4 : 1

8 : 1

7 : 1

49 : 1

From figure I1​=4I​ and I2​=649I​⇒I1​I2​​=169​ By using Imin​Imax​​=(I1​I2​​​−1I1​I2​​​+1​) =(169​​−1169​​+1​)=149​

From figure $I_{1} = \frac{I}{4}$ and $I_{2} = \frac{9 I}{64} \Rightarrow \frac{I _{2}}{I _{1}} = \frac{9}{16}$

By using $\frac{I _{m a x}}{I _{m i n}} = (\frac{\frac{I _{2}}{I _{1}} + 1}{\frac{I _{2}}{I _{1}} - 1})$
$= (\frac{\frac{9}{16} + 1}{\frac{9}{16} - 1}) = \frac{49}{1}$