Question

If transmissive power of a surface is $\frac{1}{4}$ and reflective power is $\frac{1}{6},$ then its emissivity will be

• A. $\frac{5}{12}$
• B. $\frac{12}{7}$
• C. $\frac{12}{5}$
• D. $\frac{7}{12}$

Answer 1

The correct option is D $\frac{7}{12}$

r + t + a = 1
$\frac{1}{6}+\frac{1}{4}+a=1$
$a=1-\frac{1}{4}-\frac{1}{6}$
$\frac{24}{24}-\frac{6}{24}-\frac{4}{24}=\frac{14}{24}=\frac{7}{12}$