A small bulb is placed at the bottom of a tan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The depth of the bulb in water is $h = 80 \; cm = 0.8 \; m$.The refractive index of water is $\mu = 1.33$.Light emerges from the surface only if t

Vedclass · Accepted Answer

$2.61$

Vedclass · Answer

(B) The depth of the bulb in water is $h = 80 \; cm = 0.8 \; m$.The refractive index of water is $\mu = 1.33$.Light emerges from the surface only if the angle of incidence is less than or equal to the critical angle $i_c$. At the critical angle,the angle of refraction is $r = 90^{\circ}$.Using Snell's law: $\mu \sin i_c = 1 \sin 90^{\circ} \implies \sin i_c = \frac{1}{\mu} = \frac{1}{1.33} \approx 0.7519$.From the geometry,the radius $R$ of the circular area on the surface is given by $R = h 	an i_c$.Since $\sin i_c = 0.7519$,we have $\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - (0.7519)^2} \approx 0.6593$.Thus,$	an i_c = \frac{\sin i_c}{\cos i_c} = \frac{0.7519}{0.6593} \approx 1.1404$.The radius $R = 0.8 	imes 1.1404 \approx 0.9123 \; m$.The area of the circular surface is $A = \pi R^2 = \pi 	imes (0.9123)^2 \approx 3.14159 	imes 0.8323 \approx 2.61 \; m^2$.

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \; cm$. What is the area (in $m^2$) of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

Similar Questions

If a ray of light in a denser medium strikes a rarer medium at an angle of incidence $i$,the angles of reflection and refraction are $r$ and $r^{\prime}$ respectively. If the reflected and refracted rays are at right angles to each other,the critical angle for the given pair of media is

The refractive index of glass is $\frac{3}{2}$ and that of water is $\frac{4}{3}$. The critical angle for a ray of light going from glass to water is

In the given figure,at the water-air interface,a light ray is incident at the critical angle. Find the value of $\mu_g$.

$A$ transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$. It is surrounded by air. $A$ light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

$A$ diver at a depth of $12 \ m$ in water $(\mu = 4/3)$ sees the sky in a cone of semi-vertical angle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module