JEE Challenger
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PhyMedAdv 2015 • Paper 1

Distance Between Glass Rods for Parallel Light Beam Emergence

Two identical glass rods S1S_1 and S2S_2 (refractive index =1.5= 1.5) have one convex end of radius of curvature 10 cm10\text{ cm}. They are placed with the curved surfaces at a distance dd as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light PP is placed inside rod S1S_1 on its axis at a distance of 50 cm50\text{ cm} from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2S_2. The distance dd is

Question Diagram 1

Options

A

60 cm60\text{ cm}

B

70 cm70\text{ cm}

Correct
C

80 cm80\text{ cm}

D

90 cm90\text{ cm}

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and RefractionLenses and Mirrors

Step-by-Step Solution

To find the distance dd between the curved surfaces of the two glass rods S1S_1 and S2S_2, we apply the spherical surface refraction formula:

μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}

where light travels from a medium of refractive index μ1\mu_1 to a medium of refractive index μ2\mu_2. The direction of incident light is taken as positive.

Step 1: Refraction at the curved surface of S1S_1

For the first surface (vertex V1V_1):

  • Refractive index of first medium (glass rod S1S_1), μ1=1.5\mu_1 = 1.5
  • Refractive index of second medium (air), μ2=1.0\mu_2 = 1.0
  • Object distance, u1=−50 cmu_1 = -50\text{ cm} (since point source PP is inside S1S_1)
  • The end is convex, meaning it bulges outwards to the right. Therefore, its center of curvature C1C_1 lies inside S1S_1 (to the left of V1V_1), giving R1=−10 cmR_1 = -10\text{ cm}.

Applying the refraction formula at V1V_1:

1.0v1−1.5−50=1.0−1.5−10\frac{1.0}{v_1} - \frac{1.5}{-50} = \frac{1.0 - 1.5}{-10}

1v1+3100=−0.5−10\frac{1}{v_1} + \frac{3}{100} = \frac{-0.5}{-10}

1v1+3100=120\frac{1}{v_1} + \frac{3}{100} = \frac{1}{20}

1v1=120−3100=5100−3100=2100=150\frac{1}{v_1} = \frac{1}{20} - \frac{3}{100} = \frac{5}{100} - \frac{3}{100} = \frac{2}{100} = \frac{1}{50}

v1=+50 cmv_1 = +50\text{ cm}

Thus, after refraction at V1V_1, the rays converge towards an image point I1I_1 located at a distance of 50 cm50\text{ cm} to the right of V1V_1 in air.

Step 2: Refraction at the curved surface of S2S_2

For the second surface (vertex V2V_2):

  • Refractive index of first medium (air), μ1′=1.0\mu_1' = 1.0
  • Refractive index of second medium (glass rod S2S_2), μ2′=1.5\mu_2' = 1.5
  • The final light rays inside S2S_2 emerge parallel to the principal axis, so v2=∞v_2 = \infty.
  • The curved end of S2S_2 bulges towards the left, so its center of curvature C2C_2 lies to the right of V2V_2 inside S2S_2, giving R2=+10 cmR_2 = +10\text{ cm}.

Applying the refraction formula at V2V_2:

1.5∞−1.0u2=1.5−1.0+10\frac{1.5}{\infty} - \frac{1.0}{u_2} = \frac{1.5 - 1.0}{+10}

0−1u2=0.510=1200 - \frac{1}{u_2} = \frac{0.5}{10} = \frac{1}{20}

u2=−20 cmu_2 = -20\text{ cm}

This means that the incoming rays must converge towards an object point located 20 cm20\text{ cm} to the left of V2V_2.

Step 3: Calculating distance dd

The image I1I_1 formed by S1S_1 acts as the object for S2S_2. Thus, I1I_1 must lie 20 cm20\text{ cm} to the left of V2V_2.

Since I1I_1 is 50 cm50\text{ cm} to the right of V1V_1 and 20 cm20\text{ cm} to the left of V2V_2, the separation dd between the vertices V1V_1 and V2V_2 is:

d=50 cm+20 cm=70 cmd = 50\text{ cm} + 20\text{ cm} = 70\text{ cm}

Correct Answer: (B) 70 cm70\text{ cm}

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