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Manipal Entrance Test 2014 Physics Paper
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© examsnet.com
Question : 26
Total: 60
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
×
10
−
7
m
entering the eye, with pupil area
10
−
6
m
2
, per second for vision will be nearly
100
200
300
400
Validate
Solution:
On using
I
=
P
A
;
where
P
=
radiation power
⇒
P
=
l
×
A
⇒
n
h
c
t
λ
=
I
A
⇒
n
t
=
l
A
λ
h
c
Hence, number of photons entering per sec the eye
(
n
t
)
=
10
−
10
×
10
−
6
×
5.6
×
10
−
7
6.6
×
10
−
34
×
3
×
10
8
=
300
© examsnet.com
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