Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is ____ cm.
Answer: 30
Solution
Refraction at a single spherical surface obeys vμ2−uμ1=Rμ2−μ1. With light passing from the rarer into the denser medium, μ1=1.0, μ2=1.5, the object distance u=−15cm, and R=+30cm because the centre of curvature lies on the denser side. Substituting, v1.5=300.5−151=601−604=−201, which gives v=−30cm; the negative sign shows the image is virtual and on the same side as the object. The image is formed at a distance of 30cm from the pole.
