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A thin equiconvex lens madeof glass of refractive index 3//2 and of focal length 0.3m in air is sealed into an opening at one end of a tank filled with water (mu=(3)/(2)). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis as shown in Figure. The separation between the lens and mirror is 0.8m.A small objectis placed outside the tank infront of the lens at a distance oa 0.9m from the lens along its axis. Find the position (relative to lens) of the image of the object formed by the system. |
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Answer» SOLUTION :If R is radius of curvature of each lens surface, then for air on either side of lens, we have `1/(f)=(._amu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((3)/(2)-1)((1)/(R)+(1)/(R))` `rArr (1) /(f)=(1)/(R)rArr f=RrArrR=F=0.3m` If `mu_(2)` is refractive index of lens material and `mu_(1),mu_(2)` are refractive indices on either side of the lens, then the formula is `(mu_(3))/(v)-(mu_(1))/(mu)= (mu_(2)-mu_(1))/(R_(1))+(mu_(3)-mu_(2))/(R_(2))` Here `R_(1)=0.3m, R_(2)=-0.3m, mu_(1)=mu_(air)=1` , ` mu_(2)=mu_("glass")=(3)/(2), mu_(3)=mu_("WATER")=(4)/(3), u=-9m, v=?` `:. (4//3)/(v) -(1)/((-0.9))=(((3)/(2)-1))/(0.3)+(((4)/(3)-(3)/(2)))/(-0.3)` or `(4)/(3v)+(1)/(0.9)=(1)/(0.6)+(1)/(1.8)` or `(4)/(3v)=(1)/(0.6)+(1)/(1.8)=(1)/(0.9)=(1)/(0.9)` or `v=(0.9xx4)/(3) =1.2m` That is image`I_(1)` is at a DISTANCE 1.2m from lens L to the right or at a distance `(1.2-0.8)=0.4m`to the right of mirror M. This image `I_(1)` acts as a virtual source for the plane mirror and forms real image `I_(1)` at a distance 0.4m on to the left of mirror and hence at a distance `(0.8-0.4)=0.4m` to the right of convex lens. This image `I_(2)` acts as an object for the lens. Again, the formula if `(mu_(2)^('))/(v^(''))-(mu_(1)^('))/(mu^(''))=(mu_(2)^(')-mu_(1)^('))/(R_(1))+(mu_(3)^(')-mu_(2)^('))/(R_(2))` But now `mu_(1)^(')=mu_("water")=(4)/(3),mu_(3)^(')=mu _("glass")=(3)/(2), mu_(2)^(')=mu_("air")=1` `R_(1)=-0.3m, R_(2)= +0.3m,u^('')=0.4m,v^('')=?` `:. (1)/(v^(''))-(1)/(0.3)=-(1)/(1.8)-(1)/(0.6)` or `(1)/(v^(''))=(1)/(0.3)-(1)/(1.8)-(1)/(0.6)=(1)/(0.9) rArr v^('')= 0.9m` That is image `I_(3)` is FORMED to the right at a distance 0.9m from the lens and is virtual. That is position of final iamge will be 0.9m to the right of lens. 
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