SATHEE: Optics Question 494

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 $I \times A t = \frac{L}{r^{2}} A t$

$\Rightarrow \frac{L_{1}}{r_{1}^{2}} A_{1} t_{1} = \frac{L_{2}}{r_{2}^{2}} A_{2} t_{2}$

$\Rightarrow \frac{t_{1}}{r_{1}^{2}} = \frac{t_{2}}{r_{2}^{2}}$ ( $∵ L_{1} = L_{2}$ 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.

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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

Options:

Answer:

Solution:

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