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QJEEAdv18Ph7b65: JEE Advanced 2018 Physics - Class 12 JEEAdv18Ph7b65 Solution: Sunlight of intensity is incident normally on a thin convex…

Sunlight of intensity 1.31.3 kW m−2kW\,{m^{ - 2}} is incident normally on a thin convex lens of focal length 2020 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,kW\,{m^{ - 2}}, at a distance 2222 cm from the lens on the other side is ___________.

Step-by-Step Verified Solution

Given, intensity of sunlight, I = 1.3 kW m-2 and focal length of convex lens, f = 20 cm

JEE Advanced 2018 Paper 1 Offline Physics - Geometrical Optics Question 55 English Explanation

Now, we know, Intensity = EnergyArea{{Energy} \over {Area}}

So, intensity on screen is given as

Iscreen=Incident energyA2{I_{screen}} = {{Incident\,energy} \over {{A_2}}}

But, Incident energy = Intensity of sunlight ×\times Area of lens

= 1.3 kW m −- 2 ×\times A1

So, intensity on screen is given as

Iscreen=1.3 kW m−2×A1A2{I_{screen}} = 1.3\,kW\,{m^{ - 2}} \times {{{A_1}} \over {{A_2}}}

Now from figure, and using similar triangles theorem, we have

A1A2=(202)2=100{{{A_1}} \over {{A_2}}} = {\left( {{{20} \over 2}} \right)^2} = 100

Therefore, intensity on screen = 1.3 kW m −- 2 ×\times 100 = 130 kW m −- 2

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