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PhyHardAdv 2024 • Paper 1

Image Formation in Plano Convex Beaker Filled with Liquid

A glass beaker has a solid, plano-convex base of refractive index 1.601.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is 9 cm9\text{ cm}, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index nn up to the level QPR. If the image of a point object O at a height of hh (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?

Question Diagram 1

Options

A

For n=1.42n = 1.42, h=50 cmh = 50\text{ cm}.

Correct
B

For n=1.35n = 1.35, h=36 cmh = 36\text{ cm}.

Correct
C

For n=1.45n = 1.45, h=65 cmh = 65\text{ cm}.

D

For n=1.48n = 1.48, h=85 cmh = 85\text{ cm}.

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and RefractionLenses and Mirrors

Step-by-Step Solution

To determine the condition under which the image of the point object OO is formed onto itself, we trace paraxial light rays originating from OO through the optical system:

  1. Refraction at the flat liquid surface (QPRQPR):

    • Refractive index of air: μ1=1\mu_1 = 1
    • Refractive index of liquid: μ2=n\mu_2 = n
    • Object distance from the apex PP: u1=−hu_1 = -h

    Using the refraction formula for a plane surface (R1=∞R_1 = \infty): μ2v1−μ1u1=0\frac{\mu_2}{v_1} - \frac{\mu_1}{u_1} = 0 nv1−1−h=0  ⟹  v1=−nh\frac{n}{v_1} - \frac{1}{-h} = 0 \implies v_1 = -n h

    This forms a virtual image at a distance of nhnh above PP.

  2. Refraction at the convex glass surface (SPUSPU):

    • Refractive index of liquid: μ2=n\mu_2 = n
    • Refractive index of glass base: μ3=1.60\mu_3 = 1.60
    • Radius of curvature of SPUSPU: R=+9 cmR = +9\text{ cm} (since the center of curvature lies inside the glass, along the direction of incident light)
    • Object distance for this surface: u2=v1=−nhu_2 = v_1 = -n h

    For the image to be formed onto itself, light rays entering the glass base must fall normally on the planar mirror surface STUSTU. This means the rays inside the glass must be parallel to the central optical axis (OPTOPT), which corresponds to an image distance v2=∞v_2 = \infty.

    Using the spherical surface refraction formula: μ3v2−μ2u2=μ3−μ2R\frac{\mu_3}{v_2} - \frac{\mu_2}{u_2} = \frac{\mu_3 - \mu_2}{R}

    Substituting the values: 1.60∞−n−nh=1.60−n9\frac{1.60}{\infty} - \frac{n}{-n h} = \frac{1.60 - n}{9} 0+1h=1.60−n90 + \frac{1}{h} = \frac{1.60 - n}{9} h=91.60−n cmh = \frac{9}{1.60 - n}\text{ cm}

  3. Evaluating the options:

    • For Option A (n=1.42n = 1.42): h=91.60−1.42=90.18=50 cmh = \frac{9}{1.60 - 1.42} = \frac{9}{0.18} = 50\text{ cm} (Option A is correct)

    • For Option B (n=1.35n = 1.35): h=91.60−1.35=90.25=36 cmh = \frac{9}{1.60 - 1.35} = \frac{9}{0.25} = 36\text{ cm} (Option B is correct)

    • For Option C (n=1.45n = 1.45): h=91.60−1.45=90.15=60 cm≠65 cmh = \frac{9}{1.60 - 1.45} = \frac{9}{0.15} = 60\text{ cm} \neq 65\text{ cm} (Option C is incorrect)

    • For Option D (n=1.48n = 1.48): h=91.60−1.48=90.12=75 cm≠85 cmh = \frac{9}{1.60 - 1.48} = \frac{9}{0.12} = 75\text{ cm} \neq 85\text{ cm} (Option D is incorrect)

Correct Option(s): A and B

Image Formation in Plano Convex Beaker Filled with Liquid | Physics PYQ Solution - JEE Challenger