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A kid of height 1.1 ft is sleeping straight between focus and centre of curvature along the principal axis of a concave mirror of small aperture. His head is towards the mirror and is 0.5 ft from the focus of the mirror. How a plane mirror should be placed so that the image formed by it due to reflected light from concave mirror looks like a person of height 5.5 ft standing vertically. Draw the ray diagram. Find the focal length of the concave mirror. |
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Answer» Now, for image formation of `A` and `B` by CONCAVE mirror, `u_(A)=-(F+1.6),f=-F` `u_(B)=-(F+0.5),f=-F` `v_(A)=(u_(A)f)/((U-f))=(-(F+1.6)(-F))/((-F-1.6+F))=(-(F+1.6))/(1.6)` `v_(B)=(-F(F+0.5))/(0.5)` size of the image `=|v_(B)|-|v_(A)|` `RARR 5.5=(F(F+0.5))/(0.5)-(F(F+1.6))/(1.6)` `F=2ft` The plane mirror should be placed at angleof `45^(@)` with `-ve x-` axis (as show) to get the requred vertical image `A"B"` From Newton's formula, `xy=f^(2)` so for `A,(1.1+0.5)y=f^(2)......(i)` and for `B,` `(0.5)xx(y+5.5)+f^(2) ....(ii)` from `(i)` and `(ii)` `f=2ft`. 
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