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A cube of side `2m` is placed in front of a concave mirrorr of focal length `1m` with its face `A` at a distance of `3m` and face `B` at a distance of `5m` form the mirrorr, The distance between the images of face `A` and `B` and height of images of `A` and `B` are respectively. A. `1m, 0.5,0.25m`B. `0.5m,1m,0.25m`C. `0.5m,0.25m,1n`D. `0.25m,1m,0.5m` |
Answer» Correct Answer - D For A `u=-3m` `v_(1)=?,` `f=-1m` `(1)/(u_(1))=(1)/(f)-(1)/(u)=(1)/(-1)-(1)/(-3)=(1)/(3)=1=-(2)/(3)` or `v_(1)=-(3)/(2)` For B `(1)/(v_(2))=(1)/(-1)-(1)/(-5)` or `(1)/(v_(2))=(1)/(5)-1=-(4)/(5)` or `v_(2)=-(5)/(4)m` Now, `v_(1)-v_(2)=(3)/(2)-(-(5)/(4))` `=-(3)/(2)+(5)/(4)=-(1)/(4)m=-0.25m` Again, `(I_(1))/(Q)=-(v_(1))/(u)` or `I_(1)=-(v_(1))/(u)Q-((-3)/(2))((-1)/(3))=-1m` Again `(I_(2))/(Q)=-(v_(2))/(u)` or `I_(1)=(-(5)/(4))((1)/(-5))2= -0.5m` |
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