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VITEEE 2014 Paper

Section: Physics

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Question : 40 of 120

Marks: +1, -0

An eye can detect 5 x 104 photons per square meter per sec of green light (λα = 5000 Å) while the ear can detect

10^{13}

m^{2}

. The factor by which the eye is more sensitive as a power detector then ear is close to

5
10
$10^{6}$
15

Solution:

Energy =

\frac{12375}{5000}

= 2.475 eV = 4 ×

10^{-19}

J

Minimum intensity to which the eye can respond.

I_{\text{eye}}

= (photon flux) x energy of a photon

I_{\text{eye}}

=

(5 \times 10^{4})

× 4 ×

10^{-19}

= 2 ×

10^{-14}

W/

m^{2}

Now, lesser the intensity required by a detector for detection more sensitive it will be =

\frac{I_{\text{ear}}}{I_{\text{eye}}}

=

\frac{10^{-13}}{2 \times 10^{-14}}

= 5

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