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Geometrical Optics question

2025 · 8 Apr · Shift 2 · Q53
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  5. /2025 · 8 Apr · Shift 2 · Q53

Geometrical Optics question

2025 · 8 Apr · Shift 2 · Q53

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be
  1. A
    70011\frac{700}{11}11700​ cm
  2. B
    60011\frac{600}{11}11600​ cm
  3. C
    80011\frac{800}{11}11800​ cm
  4. D
    50011\frac{500}{11}11500​ cm
View written solutionFree

Correct answer: B

  1. Interpret the system

A concave-convex glass lens has refractive index

μg=1.5\mu_g = 1.5μg​=1.5

and a liquid of refractive index

μl=1.3\mu_l = 1.3μl​=1.3

fills the upper concave surface.

So the optical system acts like:

  • one refracting spherical surface between liquid and glass,
  • and the other refracting spherical surface between glass and air.

We treat the combination by adding powers of the two surfaces.

  1. Use surface power formula

For a spherical refracting surface,

P=μ2−μ1RP = \frac{\mu_2-\mu_1}{R}P=Rμ2​−μ1​​

where light travels from medium μ1\mu_1μ1​ to medium μ2\mu_2μ2​.

Take light from top to bottom.

  1. Assign radii with sign convention

Using Cartesian sign convention (positive to the right, light downward):

  • Upper surface is concave upward. Its center of curvature lies upward, i.e. on the incident side, so
R1=−30 cmR_1 = -30\text{ cm}R1​=−30 cm
  • Lower surface is convex downward. Its center of curvature lies upward relative to that surface as well, hence
R2=−20 cmR_2 = -20\text{ cm}R2​=−20 cm
  1. Power of first surface (liquid to glass)
P1=μg−μlR1=1.5−1.3−30=0.2−30=−1150P_1 = \frac{\mu_g-\mu_l}{R_1} = \frac{1.5-1.3}{-30} = \frac{0.2}{-30} = -\frac{1}{150}P1​=R1​μg​−μl​​=−301.5−1.3​=−300.2​=−1501​
  1. Power of second surface (glass to air)
P2=1−μgR2=1−1.5−20=−0.5−20=140P_2 = \frac{1-\mu_g}{R_2} = \frac{1-1.5}{-20} = \frac{-0.5}{-20} = \frac{1}{40}P2​=R2​1−μg​​=−201−1.5​=−20−0.5​=401​
  1. Net power of combination
P=P1+P2=−1150+140P = P_1 + P_2 = -\frac{1}{150} + \frac{1}{40}P=P1​+P2​=−1501​+401​

LCM of 150 and 40 is 600:

P=−4600+15600=11600P = -\frac{4}{600} + \frac{15}{600} = \frac{11}{600}P=−6004​+60015​=60011​

Thus focal length is

f=1P=60011 cmf = \frac{1}{P} = \frac{600}{11}\text{ cm}f=P1​=11600​ cm
  1. Match with options
f=60011 cmf = \frac{600}{11}\text{ cm}f=11600​ cm

So the correct option is:

B. 60011\dfrac{600}{11}11600​ cm

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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