Here the light passes through three refracting media, I, II \& III as shown in the figure.
Media I \& II has a refractive index 1.5 and radius of curvature 20cm. While medium II has a refractive index of 1.7 and it acts as a concave lens.
Let the focal length of the three media be f1,f2,f3.
Applying the lens makers formula
1f=(μ−1)(1R1−1R2)
For media I,
R1=∞ since it is a plano convex lens.
1f1=(1.5−1)(1∞−1−20)=0.5×120=140
For media II,
1f2=(1.7−1)(1−20−120)=0.7×(−220)=−0.710
For media III,
Since media I & III are similar we can write
1f3=140
For applying the formula for equivalents lenses,
1feq=1f1+1f2+1f3
⇒1feq=140−0.710+140
⇒1feq=1−2.8+140=−0.840=−150
∴feq=−50cm
Hence the correct answer is option (D).