 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit/9/c/solution

Substituting the stated refractive index and thickness coefficient,

α = 2 m^{- 1}

, gives

f = \frac{1}{2 ( 1.5 - 1 ) ( 2 m ^{- 1} )} = 0.50 m .

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = 2 - \frac{1}{0.7} = \frac{4}{7} m^{- 1}, v = 1.75 m .

The positive image distance places a real image on the outgoing side of the optical lens, at

x = + 1.75 m

. The object is beyond the focal point, so outgoing light rays actually converge there rather than only their backward extensions meeting.

 New to topics? Read the docs here!