With respect to air,the critical angle in a m | vedclass

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

Question

(C) The critical angle $C$ is given by the formula $C = \sin^{-1}\left(\frac{1}{\mu}ight)$.For red light,$	heta = \sin^{-1}\left(\frac{1}{\mu_{\lam

Vedclass · Accepted Answer

Less than $	heta$

Vedclass · Answer

(C) The critical angle $C$ is given by the formula $C = \sin^{-1}\left(\frac{1}{\mu}ight)$.For red light,$	heta = \sin^{-1}\left(\frac{1}{\mu_{\lambda_1}}ight)$,and for yellow light,$	heta' = \sin^{-1}\left(\frac{1}{\mu_{\lambda_2}}ight)$.According to Cauchy's dispersion formula,the refractive index $\mu$ is inversely related to the wavelength $\lambda$. Since the wavelength of yellow light is less than the wavelength of red light $(\lambda_2  \mu_{\lambda_1})$.Since $\mu_{\lambda_2} > \mu_{\lambda_1}$,it follows that $\frac{1}{\mu_{\lambda_2}} Therefore,$	heta' < 	heta$. Thus,the critical angle for yellow light will be less than $	heta$.

With respect to air,the critical angle in a medium for light of red colour $[{\lambda _1}]$ is $\theta$. Other factors remaining the same,the critical angle for light of yellow colour $[{\lambda _2}]$ will be:

Similar Questions

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7} \; m$. The refractive index of water is $\frac{4}{3}$. The area of the surface of water through which light from the bulb can emerge out is $x \,\pi \; m^{2}$. The value of $x$ is ...............

For a light ray to undergo total internal reflection ($i =$ angle of incidence,$i_c =$ critical angle):

The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium into air,the angle of incidence,at which it will be totally internally reflected,is ......... $^o$

$A$ ray of light incident normally on an isosceles right-angled prism travels as shown in the figure. The least value of the refractive index of the prism must be

It is found that all electromagnetic signals sent from $A$ towards $B$ reach point $C$. The speed of electromagnetic signals in glass can not be

Vedclass Test Series

Exam Paper Generator

Online Exam Module