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Q.For refraction at any spherical surface establish the relation mu2/v - mu1/u = (mu2 - mu1)/R, where the terms have usual meanings.

Jharkhand JacJAC Intermediate Board 2025Subjective· 5mImportance★★★★★
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For a paraxial ray refracting at a single spherical surface, writing Snell's law in its small-angle form and using the exterior-angle property of the two triangles formed at the surface, then applying the Cartesian sign convention, gives mu2/v - mu1/u = (mu2-mu1)/R.

Setup: Consider a convex spherical refracting surface of radius RR with pole PP and centre of curvature CC, separating a rarer medium of refractive index μ1\mu_1 (in which the object lies) from a denser medium of index μ2\mu_2. A point object OO lies on the principal axis. A paraxial ray OMOM from OO strikes the surface at a point MM very close to the axis (foot of perpendicular NN, height NM=hNM = h), refracts, and meets the axis again at the image II.

Let the ray make the following small angles with the principal axis:

  • α=∠MOP\alpha = \angle MOP (incident ray OMOM with axis),
  • β=∠MCP\beta = \angle MCP (normal MCMC with axis),
  • γ=∠MIP\gamma = \angle MIP (refracted ray MIMI with axis),

and let ii = angle of incidence (between OMOM and normal MCMC), rr = angle of refraction (between MIMI and normal MCMC).

Exterior-angle relations: In triangle OMCOMC, the angle ii is the exterior angle at MM, so it equals the sum of the two interior opposite angles:

i=α+βi = \alpha + \beta

In triangle MCIMCI, the angle β\beta is the exterior angle at CC, so

β=r+γ    ⇒    r=β−γ\beta = r + \gamma \;\;\Rightarrow\;\; r = \beta - \gamma

Snell's law (paraxial form): For small angles sin⁡θ≈θ\sin\theta \approx \theta, so μ1sin⁡i=μ2sin⁡r\mu_1\sin i = \mu_2\sin r becomes

μ1 i=μ2 r    ⇒    μ1(α+β)=μ2(β−γ)\mu_1\, i = \mu_2\, r \;\;\Rightarrow\;\; \mu_1(\alpha + \beta) = \mu_2(\beta - \gamma)

Small-angle substitutions: Since all angles are small and NN is close to PP, using tan⁡θ≈θ\tan\theta\approx\theta:

α≈NMPO=hPO,β≈NMPC=hPC,γ≈NMPI=hPI\alpha \approx \dfrac{NM}{PO} = \dfrac{h}{PO}, \qquad \beta \approx \dfrac{NM}{PC} = \dfrac{h}{PC}, \qquad \gamma \approx \dfrac{NM}{PI} = \dfrac{h}{PI}

Substituting and dividing throughout by hh:

μ1(1PO+1PC)=μ2(1PC−1PI)\mu_1\left(\dfrac{1}{PO} + \dfrac{1}{PC}\right) = \mu_2\left(\dfrac{1}{PC} - \dfrac{1}{PI}\right)

Applying the Cartesian sign convention: distances are measured from the pole PP; those against the incident light are negative, those along it positive. Here PO=−uPO = -u, PI=+vPI = +v, PC=+RPC = +R:

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