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Qabool Jain
Asked: 2022-11-07T07:37:18+05:30 In:

A transparent solid sphere of radius `2 cm` and density `rho` floats in a transparent liquid of density `2rho` kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature `8 cm` and is silvered to make it concave mirror as shown in the figure. When an object is placed at a distance of `10 cm` directly above the centre of the sphere C, its final image coincides with it. Find h (as shown in the figure ), the height of the liquid surface in the beaker from the apex of the bottom. Consider the paraxial rays only. The refractive index of the sphere is `3//2` and that of the liquid is `4//3.`
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A transparent solid sphere of radius `2 cm` and density `rho` floats in a transparent liquid of density `2rho` kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature `8 cm` and is silvered to make it concave mirror as shown in the figure. When an object is placed at a distance of `10 cm` directly above the centre of the sphere C, its final image coincides with it. Find h (as shown in the figure ), the height of the liquid surface in the beaker from the apex of the bottom. Consider the paraxial rays only. The refractive index of the sphere is `3//2` and that of the liquid is `4//3.`
image
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  1. Correct Answer – A::C
    `(V_i)/V=(rho_S)/(rho_L)=rho/(2rho)=1/2`
    i.e. half the sphere is inside the liquid. For the
    image to coincide with the object light should fall
    normally on te sphere.
    Using `(mu_2)/v-(mu_1)/u=(mu_2-mu_1)/R` twice, we have
    `(3//2)/(v_1)-1/(-8)=(3//2-1)/(+2)`
    `:. v_1=12 cm`
    Further, `(4//3)/(h-10)-(3//2)/8=(4//3-3//2)/(-2)`
    Solving this equation, we get
    `h=15 cm`

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