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Find the radius of the light spot on the screen as shown in Fig. 34-44. Consider only the light rays refracted from the lens. Find the average intensity on the screen. The source is point source of power 100W. The lens has an aperture of 6cm. |
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Answer» Solution :(1) Only a real image can be FOCUSED on a screen. A virtual image cannot be focused on a screen but can be seen with naked eyes. To see a real image with a naked eye, we should place the eye behind the image. (2) Here, we will TRY to locate the image formed by the lens. This image will be formed by all the rays collected by the lens. So, if we consider the extreme rays which are being focused at the aperture (Fig. 35-45), we will get the FIELD of view for the image. This will also show us the region in which the refracted light will be present. Calculation : If the screen was not present, the image would have formed at `v=(uf)/(u+f)=(-30xx20)/(-30+20)=60CM` From the lens,  From similar triangles , we can estimate the radius of the image formed : `(R )/(3)=(20)/(60)` `R=1cm` The amount of light used to form the image can be estimated with the help of solid angle subtened by the lens at the location of the source : `("Power of image")/(100W)=(1-costheta)/(2)` Here, `tantheta=3//30=0.1`, for a small angle and `costheta=(1-theta^(2)//2)`. The power of the image `50xx(0.1)^(2)//2=0.25W` Intensity `=("Power")/("area")` `=(0.25)/(pixx(0.01)^(2))=(2500)/(pi)W//m^(2)`. Learn : Here we can see how the itensity can be quite high even THOUGH the power transmitted by the lens is very small . 
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