Figure shown objectO.Final image I isformed after two refraction and one refraction is also shown in figure .Findthe length of mirror.(in cm).

Figure shown objectO.Final image I isformed after two refraction and o

1.	Figure shown objectO.Final image I isformed after two refraction and one refraction is also shown in figure .Findthe length of mirror.(in cm).
Answer» `10` `15` `20` `25` SOLUTION :`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R )` for reflraction at place SURFACE `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(oo)` `(mu_(2))/(v)=(mu_(1))/(u)` ` v=(mu_(2))/(mu_(1))xxu` =`(1)/(4//3)xx(-33.25)` `v=-25cm` Image of object due to first refraction will act as object for MIRROR.Image due to reflected rays will acts as virtual object for refraction, `(mu_(2))/(v)=(mu_(1))/(u) rArr u=(mu_(1))/(mu_(2))v` ` u=(1xx25)/(4//3) rArr u=(75)/(4)CM` Applying mirror formula `(1)/(v)+(1)/(u)=(1)/(f) rArr (1)/(-(15+75//4))+(1)/(-(15+25))=(1)/(f)` `(1)/(f)=-[(4)/(135)+(1)/(40)] rArr (1)/(f)=-[(160+135)/(135xx40)]` `f=-(135xx40)/(295) rArr f=-27xx(2)/(3)` `f=-18cm`