Dual Nature Of Light Question 116
Question: The minimum intensity of light to be detected by human eye is $ {{10}^{-10}}W/m^{2} $ . The number of photons of wavelength $ 5.6\times {{10}^{-7}}m $ entering the eye, with pupil area $ {{10}^{-6}}m^{2} $ , per second for vision will be nearly
Options:
A) 100
B) 200
C) 300
D) 400
Show Answer
Answer:
Correct Answer: C
Solution:
By using $ I=\frac{P}{A}; $ where P = radiation power
Þ $ P=I\times A $
Þ $ \frac{nh,c}{t\lambda }=IA $
Þ $ \frac{n}{t}=\frac{IA\lambda }{hc} $ Hence number of photons entering per sec the eye $ ( \frac{n}{t} )=\frac{{{10}^{-10}}\times {{10}^{-6}}\times 5.6\times {{10}^{-7}}}{6.6\times {{10}^{-34}}\times 3\times 10^{8}} $ = 300.
BETA
