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Physics · 2026 · 2 marks
Ray Optics and Optical InstrumentsRefraction at Spherical Surfaces and by Lenses2 marksApplyvery short answer
CBSE 2026 · Region 4 · Set 1 · Q19
A small bulb is placed at the bottom of a tank, containing a transparent liquid of refractive index √ 2, to a depth of 1 m. Calculate the area of the…
A small bulb is placed at the bottom of a tank, containing a transparent liquid of refractive index $\displaystyle \sqrt{2}$, to a depth of $\displaystyle 1$ m. Calculate the area of the surface of the liquid through which light from the bulb emerges.
Marking-scheme solution
Light from a point source at depth h below the surface can emerge only within a cone of half-angle equal to the critical angle i_c.
Radius of the illuminated circle: r = h · tan i_c (as i = i_c).
sin i_c = $\displaystyle 1$/μ = $\displaystyle 1$/√$\displaystyle 2$ ⟹ i_c = $\displaystyle 45$°, so tan i_c = $\displaystyle 1$ and r = h.
Area = πr² = $\displaystyle 3.14$ × ($\displaystyle 1$)² = $\displaystyle 3.14$ m².
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