An object O (real ) is placed at focus of an equi-biconvex lens as shown in. figure-I . The refractive index of lens is `mu = 1.5` and the radius of curvature of either suface of lens is R . The lens is surronded by air . In each statement of column-I some changes are made to situation given above and information regarding final image formed as a result is given in column-II . The distance between lens and object is unchanged in all statements of column-I. match the statements in column-I with resulting image in column-II.

Answer

(A)-p , r (B) -r,s (C) -p (D) -r,s

Explanation

(A) when object is at focus , image is also real from equi- biconvex lens.
`because (1)/(f) = (mu-1) [(1)/(R_(1)) - (1)/(R_(2))]`
For equi biconvex ` f= (R)/(2(mu-1))…..(i) "As" mu uparrow f downarrow`
Hence final image position v decreases
hence `m= (v)/(u)` also decrease
(B) From (i) R`uparrow implies f uparrow`
Radius of curvature increases i.e., now the object is not placed at focus and it is case when object is between focus and optical centre. Hence image formation is virtual.
(C) Glass slab provides shift of the object in the incident ray direction. so for lens object position becomes between focus and optical centre.
(D) When medium is changed then relative `mu`.