TS EAMCET 03 May 2019 Shift 2 Solved Paper

The expression for energy of photons is given by,
E=‌

hc

λ

‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
Substitute the values.
E‌‌=‌

6.6×10−34×3×108

660nm

‌‌=3×10−19
The area of exposure of lamp at radius of 10m is calculated as
A‌‌=4πr2
‌‌=4π×102m2
The area of sensor is given by,
A′s‌‌=πr2
‌‌=π×(10−2)2
‌‌=π×10−4m2
The area of sensor is given by,
A′s‌‌=πr2
‌‌=π×(10−2)2
‌‌=π×10−4m2
Thus, the energy emitted by lamp in exposure time is given by,
E0‌‌=Power‌of‌lamp(p)×‌ exposure time ‌(t)
E0‌‌=300×100×10−3
‌‌=3J
The expression for the number of photons is given by,
N‌‌=‌

E0

E

×‌

AS′

A0

‌‌=‌

30

3×10−19

×‌

π×10−4

4π×102

‌‌=2.5×1013