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Use the mirror equation to show that (a) an object placed btween f and 2f a concave mirror produces a real image beyond 2f (b) a convex mirror always produces a virtual image independent of the location of the objcet. (c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. |
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Answer» Solution :(a) As `DeltasABC` and `A'B'C` are similar `therefore (AB)/(A'B')=(CB)/(CB')……..(I)` Again as `DeltasABP` and `A'B'P` are similar `(AB)/(A'B')=(PB)/(CB')……..(II)` From `(i)` and `(ii)` `(AB)/(A'B)=(PB)/(PB')…….(iii)` Measuring all distance from `P` , we have `CB=PB-PC` `CB'=PC-PB'` `therefore` from `(iii)` `(PB-PC)/(PC-PB')=(PB)/(PB')........(iv)` Using New Cartisian SIGN conventions, `PB=-u` `PC=-R` `PB'=-v` We GET from `(iv)` `(-u+R)/(-R+v)=(-u)/(-v)` or `+uR-uv=uv-vR` `uR+vR=2uv` Dividing both sides by `uvR`, we get `(1)/(v)+(1)/(u)=(2)/(R)` As `(1)/(v)+(1)/(u)=(2)/(R)=(2)/(2f)=(1)/(f)` `rArr (1)/(v)+(1)/(u)=(1)/(f)`  (b) According to sign conventions, `PB'=-u,PB'=+v` `PB=f,PC=2f` `DeltasABC` and `A'B'C` are similar `therefore (AB)/(A'B')=(CB)/(B'C)` But all DISTANCES along the principal axis should be measured from the pole of the mirror. `(AB)/(A'B')=(PC+PB)/(PC-PB')` Since the aperture of the mirror is small therefore `MP` can be regarded as a straight line. `DeltasMPF` and `A'B'F` are similar `(MP)/(A'B')=(PF)/(B'F)=(PF)/(PF-PB')`or ` (AB)/(A'B')=(f)/(f-v)........(ii)` from `(i)` and `(ii)` `(2f-u)/(2f-v)=(f)/(f-v)` `2f^(2)-2fv-uf+uv=2f^(2)-VF` `-fv-uf+uv=0` `uv=fv+fu` Dividing by `uvf`, we get `(1)/(f)=(1)/(u)+(1)/(v)`  (c) `(PB-PC)/(PC-PB')=(PB)/(PB') ........(i)` Proceeding as above, we get `(CB)/(CB')=(PB)/(PB')...........(ii)` Measuring all distance from `O`, we get `CB=PC-PB` `CB'=PC+PB` From `(i)` `(PC-PB)/(PC+PB') =(PB)/(PB')` Using New cartisian sign conventins, PB=-u,PB=+v` `PC=-R=(-R+u)/(-R+V)=(-u)/(v)` or `uR-uv=vR+uv` `uR+vR=2 uv` Dividing both sides by `uvR`, we get `(1)/(v)+(1)/(u)=(2)/(R)=(2)/(2f)=(1)/(f),(1)/(v)+(1)/(u)=(1)/(f)` 
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