Question 122: Ray Optics

Question

A spherical surface of radius of curvature $R$, separates air from glass (refractive index $=1.5$ ). The centre of curvature is in the glass medium. A point object ' $O$ ' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at $P$ and $P O=P I$. The distance $P O$ equals to

Accepted Answer

We wish to find the distance from the point where the ray joining the object (O) and its image (I) meets the spherical surface (P) to the object. The given condition is that
Ã¢ÂÂÃ¢ÂÂ$$PO = PI,$$
meaning that P is the midpoint of OI.
LetÃ¢ÂÂs work through the steps:
Setup the coordinate system and geometry:
Ã¢ÂÂÃ¢ÂÂ¢ Choose the vertex of the spherical surface (the Ã¢ÂÂpoleÃ¢ÂÂ) as the origin (x = 0).
Ã¢ÂÂÃ¢ÂÂ¢ Let the optical axis be the xÃ¢ÂÂaxis.
Ã¢ÂÂÃ¢ÂÂ¢ Since the object is in air, place O at x = Ã¢ÂÂu (with u > 0).
Ã¢ÂÂÃ¢ÂÂ¢ Since the real image is formed inside the glass, place I at x = v (with v > 0).
Ã¢ÂÂÃ¢ÂÂ¢ The spherical surface has radius of curvature $$R$$ and its center of curvature is in the glass. Thus, we take the center of the sphere as $$C=(R,0)$$.
Ã¢ÂÂÃ¢ÂÂ¢ The vertex (the point on the sphere closest to the object) is then located at x = 0.
Position of P:
Ã¢ÂÂÃ¢ÂÂ¢ The ray joining O and I is along the xÃ¢ÂÂaxis. Its midpoint is
Ã¢ÂÂÃ¢ÂÂ$$x_P = \frac{-u + v}{2}.$$
Ã¢ÂÂÃ¢ÂÂ¢ However, since the physical part of the spherical surface is the convex cap met first by the incident light, the ray in fact meets the surface at the vertex, which is at x = 0.
Ã¢ÂÂÃ¢ÂÂ¢ Equating the midpoint to 0, we have
Ã¢ÂÂÃ¢ÂÂ$$\frac{-u + v}{2}=0 \quad \Longrightarrow \quad v = u.$$
Ã¢ÂÂÃ¢ÂÂ¢ Thus, the object and its image are equidistant from the vertex, and
Ã¢ÂÂÃ¢ÂÂ$$PO = |0 - (-u)| = u.$$
Relate u and R using the refraction at a spherical surface:
Ã¢ÂÂÃ¢ÂÂ¢ The formula for refraction at a spherical surface (using the appropriate signÃ¢ÂÂconvention) is
Ã¢ÂÂÃ¢ÂÂ$$\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R},$$
Ã¢ÂÂÃ¢ÂÂwhere for our problem
Ã¢ÂÂÃ¢ÂÂÃ¢ÂÂÃ¢ÂÂ $$n_1 = 1$$ (air)
Ã¢ÂÂÃ¢ÂÂÃ¢ÂÂÃ¢ÂÂ $$n_2 = 1.5$$ (glass).
Ã¢ÂÂÃ¢ÂÂ¢ With $$v = u$$, substitute in:
Ã¢ÂÂÃ¢ÂÂ$$\frac{1}{u} + \frac{1.5}{u} = \frac{0.5}{R} \quad \Longrightarrow \quad \frac{2.5}{u} = \frac{0.5}{R}.$$
Ã¢ÂÂÃ¢ÂÂ¢ Solving for u:
Ã¢ÂÂÃ¢ÂÂ$$u = \frac{2.5\,R}{0.5} = 5R.$$
Conclusion:
Ã¢ÂÂÃ¢ÂÂ¢ Since $$PO = u$$, we have
Ã¢ÂÂÃ¢ÂÂ$$PO = 5R.$$
Thus, the distance $$PO$$ equals $$5R$$, which is Option A.

Answer

5R

Answer

2R

Answer

1.5R

Answer

3R

Ray Optics

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