The radius of curvature of each surface of an equiconvex lens is  R=42cm.  Refractive index of the glass =1.25 .  If the final image forms after four internal reflections (But not TIR) inside the lens (for parallel  incident beam) calculate the distance of the image from the lens.

Using μ2v-μ1u=μ2-μ1RFor first refraction, we have  μvi-1∞=μ-1R⇒vi=μμ-1R For first reflection, let us use mirror formula  1v+1u=1f ⇒1v1+μ-1μR=-2R ⇒1v1=-3μ-1μR For second reflection,  1v2+3μ-1μR=-2R⇒1v2=-5μ-1μR Similarly, after nth reflection, 1vn=-(2n+1)μ-1μR Finally, using μ2v-μ1u=μ2-μ1R, we have 1vf-(2n+1)μ-1μRμ=1-μ-R∵μ1=μ and μ2=1⇒vf=R2(μn+μ-1)In our case , internal reflections=n=4 ⇒vf=422(1.25×4+1.25-1)=4cm