In a Young's double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength \lambda is used. Then the value of \lambda is (in nm) \underline\hspace2cm .

Wave Optics question

2020 · 9 Jan · Shift 2 · Q43

2020 · 9 Jan · Shift 2 · Q43

View written solutionFree

Correct answer: 750

\beta = \frac{\lambda D}{d}

where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is slit separation.

For the same small portion of the screen, the total covered length is fixed. If $N$ fringes are observed in that same length, then

N \propto \frac{1}{\beta}

Since $\beta \propto \lambda$ , we get

N \propto \frac{1}{\lambda}

\frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1}

Here,

So,

\frac{15}{10} = \frac{\lambda}{500}

\lambda = 500 \times \frac{15}{10} = 750\,\text{nm}

Hence, the required wavelength is

\boxed{750}

2020 · 9 Jan · Shift 2 · Q43

2020 · 9 Jan · Shift 2 · Q43

View written solutionFree

Correct answer: 750

\beta = \frac{\lambda D}{d}

where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is slit separation.

For the same small portion of the screen, the total covered length is fixed. If $N$ fringes are observed in that same length, then

N \propto \frac{1}{\beta}

Since $\beta \propto \lambda$ , we get

N \propto \frac{1}{\lambda}

\frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1}

Here,

So,

\frac{15}{10} = \frac{\lambda}{500}

\lambda = 500 \times \frac{15}{10} = 750\,\text{nm}

Hence, the required wavelength is

\boxed{750}