Question

Radiant energy is incident on a body at the rate of 2000 joules per minute. If the reflection coefficient of the body is 0.1 and its transmission coefficient is 0.2, find the radiant energy 

(i) absorbed 

(ii) reflected 

(iii) transmitted by the body in 2 minutes.

Answer 1

Data : r = 0.1, t = 0.2 

a + r + t = 1 

∴ a = 1 – r – f = 1 – 0.1 – 0.2 = 0.7

Radiant energy incident per minute on the body is 2000 J. Hence, in 2 minutes, the radiant energy (Q) incident on the body is 4000 J. Let Q_a , Q_r and Q_t be the quantities of radiant energy absorbed, reflected and transmitted in 2 minutes by the body, respectively.

(i) a = \(\frac{Q_a}{Q}\) 

\(\therefore\) Q_a = aQ = 0.7 x 4000 = 2800J

(ii) r = \(\frac{Q_r}{Q}\)

 \(\therefore\) Q_r = rQ = 0.1 x 4000 = 400J

(iii) t = \(\frac{Q_t}{Q}\)

 \(\therefore\) Q_t = tQ = 0.2 x 4000 = 800J