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Geometrical Optics question

2012 · Shift 1 · Q47
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Geometrical Optics question

2012 · Shift 1 · Q47

JEE AdvancedPhysicsGeometrical OpticsMCQ+3 / −1
A biconvex lens is formed with two planoconvex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surface are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be IIT-JEE 2012 Paper 1 Offline Physics - Geometrical Optics Question 33 English
  1. A
    −-− 280.0 cm
  2. B
    40.0 cm
  3. C
    21.5 cm
  4. D
    13.3 cm
View written solutionFree

Correct answer: B

  1. Interpret the combination

The given biconvex lens is made by joining two plano-convex lenses of different refractive indices:

  • Left part: n1=1.5n_1 = 1.5n1​=1.5
  • Right part: n2=1.2n_2 = 1.2n2​=1.2
  • Both curved surfaces have radius R=14 cmR = 14\,\text{cm}R=14cm

The plane surfaces are in contact, so the system behaves like a thin combination of refracting surfaces.

  1. Find power of each plano-convex lens

For a thin plano-convex lens in air,

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

For a plano-convex lens, one surface is plane, so one curvature is zero. Hence magnitude of power is

P=n−1RP = \frac{n-1}{R}P=Rn−1​

Now, both lenses are arranged so that together they form a biconvex lens. Thus their powers add:

Peq=P1+P2P_{\text{eq}} = P_1 + P_2Peq​=P1​+P2​

For the first lens:

P1=1.5−114=0.514P_1 = \frac{1.5-1}{14} = \frac{0.5}{14}P1​=141.5−1​=140.5​

For the second lens:

P2=1.2−114=0.214P_2 = \frac{1.2-1}{14} = \frac{0.2}{14}P2​=141.2−1​=140.2​

Therefore,

Peq=0.5+0.214=0.714=0.05 cm−1P_{\text{eq}} = \frac{0.5+0.2}{14} = \frac{0.7}{14} = 0.05\,\text{cm}^{-1}Peq​=140.5+0.2​=140.7​=0.05cm−1

So the equivalent focal length is

f=1Peq=10.05=20 cmf = \frac{1}{P_{\text{eq}}} = \frac{1}{0.05} = 20\,\text{cm}f=Peq​1​=0.051​=20cm
  1. Use lens formula

Using Cartesian sign convention:

  • Object distance u=−40 cmu = -40\,\text{cm}u=−40cm
  • Focal length f=+20 cmf = +20\,\text{cm}f=+20cm

Lens formula:

1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}f1​=v1​−u1​

Substitute values:

120=1v−(−140)\frac{1}{20} = \frac{1}{v} - \left(-\frac{1}{40}\right)201​=v1​−(−401​) 120=1v+140\frac{1}{20} = \frac{1}{v} + \frac{1}{40}201​=v1​+401​

So,

1v=120−140=2−140=140\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}v1​=201​−401​=402−1​=401​

Hence,

v=40 cmv = 40\,\text{cm}v=40cm
  1. Match with options

The image distance is

40.0 cm40.0\,\text{cm}40.0cm

So the correct option is: B

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.

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