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A fly F is sitting an aglass S 45 cm thick & of refractive index 3//2. The slab covers the top of a container C containing water (R.I.4//3) upto a height of 20 cm. Bottom of container is closed by a concave mirror M fo radius of curvature 40 cm. Lacate the final image formed by all refractions & reflection assuming paraxial rays. |
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Answer» `(mu_(2))/(V)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` `(4)/(3v)-((3)/(2))/((-45))=0 (therefore R =infty) , therefore v=-40 cm` For reflection from mirror, `u=-(20 + 40) cm =- 60 cm. , f=-20 cm` `therefore (1)/(v)=(1)/(f)-(1)/(u)` gives `v=-30 cm` Again, for refraction from surface `-2` `u=30 - 20 =10 cm`. `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` gives `(mu_(s))/(v)-(mu_(W))/(u)=(mu_(s)-mu_(w))/(R)` `(3)/(2v)-(4)/(3xx(+10))=0 , v=(45)/(4) cm` For refraction from surface `-1` `u=(45-(45)/(4))cm=(3xx45)/(4)` `(mu_(a))/(v)-(mu_(s))/(u)=0`. (thereforeR=infty) `therefore v=(-45)/(2)=-22.5 cm` Hence, iamge is FORMED `22.5 cm` below surface `1` 
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