Optics Question 494
Question: To prepare a print the time taken is 5 sec due to lamp of 60 watt at 0.25 m distance. If the distance is increased to 40 cm then what is the time taken to prepare the similar print
[CPMT 1982]
Options:
A) 3.1 sec
B) 1 sec
C) 12.8 sec
D) 16 sec
Show Answer
Answer:
Correct Answer: C
Solution:
To develop a print a fix amount of energy is required. Total light energy incident on photo print $ I\times At=\frac{L}{r^{2}}At $
$ \Rightarrow \frac{L _{1}}{r _{1}^{2}}A _{1}t _{1}=\frac{L _{2}}{r _{2}^{2}}A _{2}t _{2} $
$ \Rightarrow \frac{t _{1}}{r _{1}^{2}}=\frac{t _{2}}{r _{2}^{2}} $ ( $ \because \ L _{1}=L _{2} $ and $ A _{1}=A _{2} $ )
$ \Rightarrow t _{2}=\frac{r _{2}^{2}}{r _{1}^{2}}.t _{1} $
$ =( \frac{0.40}{0.25} )\ 2\times 5 $ = 12.8 sec.
BETA
