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$
$\Rightarrow \frac{\lambda_1}{\lambda_2} = \frac{n_2}{n_1} \Rightarrow \, \, \frac{6000}{\lambda_2} = \frac{24}{16}$
$\Rightarrow \lambda_2 = \frac{6000 \times 16}{24} = \frac{96000}{24} = 4000 \mathring {A}$