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.

Question

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.

Infinity Learn · Accepted Answer

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

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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.

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