SATHEE: ray-optics-and-optical-instruments Question 32

ray-optics-and-optical-instruments Question 32

Question: Q. 7. A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of $80 cm$ . Find out the area of the surface of water through which light from the bulb can emerge. Take the value of the refractive index of water to be $\frac{4}{3}$ . A [Delhi Comptt. I, II, III 2013]

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Solution:

Ans. Actual depth of the bulb in water, $(d_{1}) = 0.8 m$

Refractive index of water, $μ = \frac{4}{3}$

Where, Angle of incidence $= θ$

Angle of refraction $= 90^{\circ}$

Since the bulb is a point source, the emergent light can be considered as a circle of radius $r$

$\begin{aligned} μ & = \frac{\sin 90^{\circ}}{\sin i} \frac{4}{3} & = \frac{\sin 90^{\circ}}{\sin θ} \sin θ & = \frac{3}{4} \cos θ & = \sqrt{1 - \sin^{2} θ} = \frac{\sqrt{7}}{4} \tan θ & = \frac{3}{\sqrt{7}} \end{aligned}$

From the figure $\tan θ = \frac{r}{h}$

$\frac{3}{\sqrt{7}} = \frac{r}{0.8}$

$∴$

Area of the surface of water through light emerge $= π r^{2} = π (0.91)^{2} = 2.6 m^{2}$

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately $2.6 m^{2}$ .

[A] Q. 8. Two monochromatic rays of light are incident normally on the face $A B$ of an isosceles rightangled prism $A B C$ . The refractive indices of the glass prism for the two rays ’ 1 ’ and ’ 2 ’ are respectively 1.35 and 1.45 . Trace the path of these rays after entering through the prism.