Question

When a light wave of frequency $5\times {10}^{14}Hz$ is passed through a medium of refractive index $1.5$ then its wavelength will become:

• A. $4000\stackrel{o}{A}$
• B. $4500\stackrel{o}{A}$
• C. $6000\stackrel{o}{A}$
• D. $9000\stackrel{o}{A}$

Answer 1

Answer: (A)

$4000\stackrel{o}{A}$

In the new medium, frequency remains the same but speed is:

$v=\frac{c}{n}=f{\lambda }^{\prime }$

${\lambda }^{\prime }=\frac{c}{nf}=\frac{3\times {10}^8}{1.5\times 5\times {10}^{14}}$

${\lambda }^{\prime }=4\times {10}^{-7}=4000\mathring{A}$