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Geometrical Optics question

2014 · Shift 1 · Q54
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Geometrical Optics question

2014 · Shift 1 · Q54

JEE AdvancedPhysicsGeometrical OpticsMultiple correct+3 / −1
JEE Advanced 2014 Paper 1 Offline Physics - Geometrical Optics Question 38 English A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then
  1. A
    ∣f1∣=3R\left| {{f_1}} \right| = 3R∣f1​∣=3R
  2. B
    ∣f1∣=2.8R\left| {{f_1}} \right| = 2.8R∣f1​∣=2.8R
  3. C
    ∣f2∣=2R\left| {{f_2}} \right| = 2R∣f2​∣=2R
  4. D
    ∣f2∣=1.4R\left| {{f_2}} \right| = 1.4R∣f2​∣=1.4R
View written solutionFree

Correct answer: A, C

  1. Key idea: a thin film on a spherical surface behaves like two refracting spherical surfaces in contact
    The film has refractive index n1=1.4n_1=1.4n1​=1.4, the cylinder has refractive index n2=1.5n_2=1.5n2​=1.5, and outside medium is air (n=1)(n=1)(n=1).

Because the film is thin and of uniform thickness, the two spherical interfaces are essentially concentric and have nearly the same radius RRR. Hence we can add the powers of the two refracting surfaces.

For a spherical refracting surface,

P=nright−nleftRP=\frac{n_{\text{right}}-n_{\text{left}}}{R}P=Rnright​−nleft​​

with Cartesian sign convention. Here the surface is convex as seen from air, and the center lies inside the cylinder, so R>0.R>0.R>0.


  1. Case 1: Rays go from air →\to→ film →\to→ glass
    Parallel rays are incident from air.
  • First surface: air (1)(1)(1) to film (1.4)(1.4)(1.4)
P1=1.4−1R=0.4RP_1=\frac{1.4-1}{R}=\frac{0.4}{R}P1​=R1.4−1​=R0.4​
  • Second surface: film (1.4)(1.4)(1.4) to glass (1.5)(1.5)(1.5)
P2=1.5−1.4R=0.1RP_2=\frac{1.5-1.4}{R}=\frac{0.1}{R}P2​=R1.5−1.4​=R0.1​

So total power is

P=P1+P2=0.4R+0.1R=0.5RP=P_1+P_2=\frac{0.4}{R}+\frac{0.1}{R}=\frac{0.5}{R}P=P1​+P2​=R0.4​+R0.1​=R0.5​

This is equivalent to a single refracting surface from air to glass:

P=1.5−1R=0.5RP=\frac{1.5-1}{R}=\frac{0.5}{R}P=R1.5−1​=R0.5​

For parallel rays from air, image is formed in glass at distance f1f_1f1​ from the surface. For a refracting surface,

n2v−n0u=n2−n0R\frac{n_2}{v}-\frac{n_0}{u}=\frac{n_2-n_0}{R}vn2​​−un0​​=Rn2​−n0​​

Here u=∞u=\inftyu=∞, n0=1n_0=1n0​=1, n2=1.5n_2=1.5n2​=1.5, so

1.5f1=0.5R\frac{1.5}{f_1}=\frac{0.5}{R}f1​1.5​=R0.5​ f1=1.5R0.5=3Rf_1=\frac{1.5R}{0.5}=3Rf1​=0.51.5R​=3R

Hence,

∣f1∣=3R|f_1|=3R∣f1​∣=3R

So A is correct and B is incorrect.


  1. Case 2: Rays go from glass →\to→ film →\to→ air
    Now parallel rays are incident from inside the glass toward air.

Again add powers:

  • First surface: glass (1.5)(1.5)(1.5) to film (1.4)(1.4)(1.4)
P1=1.4−1.5R=−0.1RP_1=\frac{1.4-1.5}{R}=\frac{-0.1}{R}P1​=R1.4−1.5​=R−0.1​
  • Second surface: film (1.4)(1.4)(1.4) to air (1)(1)(1)
P2=1−1.4R=−0.4RP_2=\frac{1-1.4}{R}=\frac{-0.4}{R}P2​=R1−1.4​=R−0.4​

Thus total power:

P=−0.1R+−0.4R=−0.5RP=\frac{-0.1}{R}+\frac{-0.4}{R}=\frac{-0.5}{R}P=R−0.1​+R−0.4​=R−0.5​

Equivalent to single refracting surface from glass to air:

P=1−1.5R=−0.5RP=\frac{1-1.5}{R}=\frac{-0.5}{R}P=R1−1.5​=R−0.5​

For parallel rays inside glass, using refraction formula:

1f2−0=1−1.5R\frac{1}{f_2}-0=\frac{1-1.5}{R}f2​1​−0=R1−1.5​

would be incorrect because object space is glass. Correctly,

1v−1.5u=1−1.5R\frac{1}{v}-\frac{1.5}{u}=\frac{1-1.5}{R}v1​−u1.5​=R1−1.5​

with u=∞u=\inftyu=∞:

1f2=−0.5R\frac{1}{f_2}=\frac{-0.5}{R}f2​1​=R−0.5​ f2=−2Rf_2=-2Rf2​=−2R

Therefore,

∣f2∣=2R|f_2|=2R∣f2​∣=2R

So C is correct and D is incorrect.


  1. Option check
  • A: ∣f1∣=3R|f_1|=3R∣f1​∣=3R ✅
  • B: ∣f1∣=2.8R|f_1|=2.8R∣f1​∣=2.8R ❌
  • C: ∣f2∣=2R|f_2|=2R∣f2​∣=2R ✅
  • D: ∣f2∣=1.4R|f_2|=1.4R∣f2​∣=1.4R ❌

  1. Final answer The correct options are:
A, C\boxed{A,\ C}A, C​
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