Question

Young's experiment is performed with light of wavelength 6000 $\mathring{A}$ where in 16 fringes occupy a certain region on the screen. If 24 fringes occupy the same region with another light, of wavelength $\lambda$, then $\lambda$ is

• A. 6000 $\mathring{A}$
• B. 4500 $\mathring{A}$
• C. 5000 $\mathring{A}$
• D. 4000 $\mathring{A}$
• E. 5500 $\mathring{A}$

Answer 1

Answer: (D)

${\lambda }_1=6000\mathring{A},n_1=16$ fringes
$\therefore n_2=24$ fringes
$v=n\lambda$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{n_2}{n_1}\Rightarrow \frac{6000}{{\lambda }_2}=\frac{24}{16}$
$\Rightarrow {\lambda }_2=\frac{6000\times 16}{24}=\frac{96000}{24}=4000\mathring{A}$