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An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the morror equation 1/f=1/u+1/v. (b) An object is placed 30 cm is front of a plano-convex lens with its spherical surface of radius of curvature 20cm. If the refractive index of the material of the lens 1.5, find the position and nature of the image formed. |
Answer» Solution :(a) Let a linear object AB be placed normally on the principal, axis of a CONCAVE mirror between its pole and principal focus so that its virtual and erect image A.B. is formed behind the mirror as shown here. Since aperture of mirror is small the arc MP MAY be considered to be a straight chord. Since `triangleCAB and triangleCA.B.` are similar triangles, we have  `(A.B.)/(AB)=(CB.)/(CB) ............(i)` Again `triangleFA.B. and triangleFM` are similar triangles, we have `(A.B.)/(MP)=(B.F)/(PF) .......(ii)` Since MP is equal to AB, equation (ii) may be written as `(A.B.)/(AB)=(B.F)/(PF) ......(iii)` Comparing (i) and (iii), we get `(CB.)/(CB)=(B.F)/(PF) or (CP+PB.)/(CP-PB)=(PB.+PF)/(PF)` Taking cartesisan sign convention, we can write that `CP=R=-2f, PF=-f, PB=-u and PB.=+V`, Thus, we have `((-2f) +(+v))/((-2f)-(-u))=((+v)+(-f))/((-f))` `rArr 2f^(2)-vf=-2fv+2f^(2)+uv-uf` `rArr vf+uf=uv` Dividing throughtout by uvf, we get `1/u+1/v=1/f` (b) Hert, u=-30 cm, `R_(1)=+20cm, R_(2)=oo` and refractive index of lens n=15 `1/f =(n-1) (1/R_(1)-1/R_(2))=(1.5-1) (1/(20) -1/oo) =1/(40)` `f=40cm` Now, as PER relation `1/v-1/u=1/f`, we have `1/v=1/u+1/f=1/(-30)+1/(40)=-(1)/(120)` The -ve sign of v shows that the image is virtual, errect and on same side of lens as the object is. |
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