Question 39: Ray Optics

Question

Sunlight of intensity $$1.3$$ $$kW\,{m^{ - 2}}$$ is incident normally on a thin convex lens of focal length $$20$$ cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in $$kW\,{m^{ - 2}},$$ at a distance $$22$$ cm from the lens on the other side is ___________.

Accepted Answer

Given, intensity of sunlight, I = 1.3 kW m
-2
and focal length of convex lens, f = 20 cm
Now, we know, Intensity = $${{Energy} \over {Area}}$$
So, intensity on screen is given as
$${I_{screen}} = {{Incident\,energy} \over {{A_2}}}$$
But, Incident energy = Intensity of sunlight $$	imes$$ Area of lens
= 1.3 kW m
$$-$$2
$$	imes$$ A
1
So, intensity on screen is given as
$${I_{screen}} = 1.3\,kW\,{m^{ - 2}} 	imes {{{A_1}} \over {{A_2}}}$$
Now from figure, and using similar triangles theorem, we have
$${{{A_1}} \over {{A_2}}} = {\left( {{{20} \over 2}} ight)^2} = 100$$
Therefore, intensity on screen = 1.3 kW m
$$-$$2
$$	imes$$ 100 = 130 kW m
$$-$$2

Ray Optics

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