JEE Challenger
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PhyMedAdv 2012 • Paper 1

Image Distance Formed by Combination of Two Lenses

A composite bi-convex lens is constructed using two thin plano-convex lenses as illustrated in the figure. The refractive index nn of the first lens is 1.51.5 and that of the second lens is 1.21.2. Both curved surfaces have an identical radius of curvature R=14 cmR = 14\text{ cm}. For an object positioned at a distance of 40 cm40\text{ cm} from this bi-convex lens, the resulting image distance is:

Question Diagram 1

Options

A

−280.0 cm-280.0\text{ cm}

B

40.0 cm40.0\text{ cm}

Correct
C

21.5 cm21.5\text{ cm}

D

13.3 cm13.3\text{ cm}

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and MirrorsReflection and Refraction

Step-by-Step Solution

To find the image distance formed by the combination of the two thin plano-convex lenses, we first calculate the focal length of each individual lens using the Lens Maker's formula:

1f=(n−1)(1R1−1R2)\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

1. Focal length of the first lens (f1f_1):

For the first lens:

  • Refractive index, n1=1.5n_1 = 1.5
  • Radius of curvature of the first (curved) surface, R1=+14 cmR_1 = +14\text{ cm}
  • Radius of curvature of the second (flat) surface, R2=∞R_2 = \infty

1f1=(1.5−1)(114−1∞)=0.5×114=128 cm−1\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{14} - \frac{1}{\infty} \right) = 0.5 \times \frac{1}{14} = \frac{1}{28}\text{ cm}^{-1}

2. Focal length of the second lens (f2f_2):

For the second lens:

  • Refractive index, n2=1.2n_2 = 1.2
  • Radius of curvature of the first (flat) surface, R1′=∞R_1' = \infty
  • Radius of curvature of the second (curved) surface, R2′=−14 cmR_2' = -14\text{ cm}

1f2=(1.2−1)(1∞−(−114))=0.2×114=170 cm−1\frac{1}{f_2} = (1.2 - 1) \left( \frac{1}{\infty} - \left(-\frac{1}{14}\right) \right) = 0.2 \times \frac{1}{14} = \frac{1}{70}\text{ cm}^{-1}

3. Equivalent focal length of the combination (FF):

Since the two thin lenses are placed in contact, their equivalent power is the sum of their individual powers:

1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} 1F=128+170\frac{1}{F} = \frac{1}{28} + \frac{1}{70}

Taking the least common multiple of 2828 and 7070, which is 140140:

1F=5+2140=7140=120 cm−1\frac{1}{F} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20}\text{ cm}^{-1}

Thus, the equivalent focal length of the combination is: F=+20 cmF = +20\text{ cm}

4. Determination of the Image Distance (vv):

Using the thin lens formula:

1v−1u=1F\frac{1}{v} - \frac{1}{u} = \frac{1}{F}

Given:

  • Object distance, u=−40 cmu = -40\text{ cm}
  • Equivalent focal length, F=+20 cmF = +20\text{ cm}

Substituting these values into the formula:

1v−1−40=120\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} 1v+140=120\frac{1}{v} + \frac{1}{40} = \frac{1}{20} 1v=120−140=140\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} v=+40.0 cmv = +40.0\text{ cm}

Hence, the resulting image distance is 40.0 cm40.0\text{ cm}.

Correct Option: B

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