Question 111: Waves Optics

Question

Two coherent sources of light interfere. The intensity ratio of two sources is $$1: 4$$. For this interference pattern if the value of $$\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$$ is equal to $$\frac{2 \alpha+1}{\beta+3}$$, then $$\frac{\alpha}{\beta}$$ will be :

Accepted Answer

$${I_{\max }} = {\left( {\sqrt {{I_1}}  + \sqrt {{I_2}} } ight)^2}$$
$${I_{\min }} = {\left( {\sqrt {{I_1}}  - \sqrt {{I_2}} } ight)^2}$$
$$	herefore$$ $${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{2({I_1} + {I_2})} \over {4 	imes \sqrt {{I_1}{I_2}} }}$$
$$ = {1 \over 2} 	imes {{\left( {{{{I_1}} \over {{I_2}}} + 1} ight)} \over {\sqrt {{{{I_1}} \over {{I_2}}}} }}$$
$$ = {1 \over 2} 	imes {{\left( {{1 \over 4} + 1} ight)} \over {\left( {{1 \over 2}} ight)}}$$
$$ = {5 \over 4} = {{2 	imes 2 + 1} \over {1 + 3}}$$
$$	herefore$$ $${\alpha  \over \beta } = {2 \over 1} = 2$$

Answer

1.5

Answer

2

Answer

0.5

Answer

1

Waves Optics

Two coherent sources of light interfere. The intensity ratio of two sources is $1: 4. For this interference pattern if the value of \frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} is equal to \frac{2 \alpha+1}{\beta+3}, then \frac{\alpha}{\beta}$ will be :

Choose the correct answer: