Question

Air has refractive index $1.0003$. Find the thickness of air column which will contain one more wavelength of yellow light of $6000$ $A^0$ than in same thickness of vacuum.

• A. $1$ mm
• B. $2$ mm
• C. $3$ mm
• D. $5$ mm

Answer 1

The correct option is B $2$ mm

$f_1=v{\lambda }_1$ Here, $v$ is the velocity of yellow light in air ${\lambda }_1$ is a wavelength of the yellow light in air

A frequency of yellow light in vacuum:

$f_2=c{\lambda }_2$ Here, $c$ is velocity of yellow light in vacuum ${\lambda }_2$ is wavelength of the yellow light in vacuum

Equate frequencies in air and vacuum:

$v{\lambda }_1=c{\lambda }_{2}cv={\lambda }_2{\lambda }_{1}But,cv=\mu a$ Here, $\mu a$ is refractive index of air $\mu a={\lambda }_2{\lambda }_1$ So, ${\lambda }_1={\lambda }_2\mu a$

Number of wavelengths in each medium:

In air, $n_1=t{\lambda }_1$

In vacuum, $n_2=t{\lambda }_2$

Here, $t$ is the thickness of each column

Difference of thickness of each column:

$n_1-n_2=t{\lambda }_1-t{\lambda }_2=t(1{\lambda }_1-1{\lambda }_2)=t{\lambda }_2({\lambda }_2{\lambda }_1-1)=t{\lambda }_2(\mu a-1)=$

t = ${\lambda }_2\mu a-1=6000\times 10-8cm1.0003-1=0.2cm=2mm$