A sphere of refractive index μ=2 has inner radius 10 cm and outer radius 20 cm. It is placed in air as shown in the figure. The distance through which point mark P appear to shift is ……. cm.

For refraction through the first surface  μ1=2,μ2=1,u=−10cm,r=+10cmAccording to refractive surface formula,∵μ2v−μ1u=μ2−μ1r ⇒1v−2−10=1−210 ⇒1v=−110−210=−1−210=−310 ∵v=−103cmThe image P1 formed by refraction through first surface behaves as object for second surface.            For second surface,  μ1'=1,μ2'=2u'=−(20+103)=−(703)cm=−703cm,r'=−10cm  ∵μ2'v'−μ1'u'=μ2'−μ1'r' ⇒2v'−1−703=2−1−10 ∴v'=−14 cm For refraction through third surface ,  μ1'=2,μ2"=1,μ"=−(14+10)=−24 cm ∵r"=−20 cm ∵μ2"v"−μ1"u"=μ2"−μ1"r" ⇒1v"−2−24=1−2−20 ⇒v"=−30 cm Thus, shifting in object = 40−30 = 10 cm