Taking the pole P as the origin and the direction of incident light as positive:
μ1=1 (air), μ2=1.5 (lens)
u=−40 cm
R=−20 cm (centre of curvature lies to the left of P)
ho=+2 cm
Using the formula for refraction at a single spherical surface:
vμ2−uμ1=Rμ2−μ1
v1.5−−401=−201.5−1
v1.5+401=−401
v1.5=−402=−201
v=−30 cm
The negative sign indicates a virtual image formed 30 cm to the left of P.
The transverse magnification for refraction at a spherical surface is:
m=μ2⋅uμ1⋅v=1.5⋅(−40)1⋅(−30)=−60−30=0.5
Therefore:
hi=m⋅ho=0.5×2=1 cm
Hence, the correct option is (1) 1.