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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics4 Marks
Question
In a biprism experiment, the distance between the second and tenth dark bands on the same side of the central bright band is $0.12 \ cm$, that between the slit and the biprism is $20 \ cm$ and that between the biprism and the eyepiece is $80 \ cm$. If the slit images given by the lens in the two positions are $4.5\ mm$ and $2\ mm$ apart, find the wavelength of light used.
✓

Answer

The distance between the second and tenth dark bands on the same side of the central band is equal to $8$ times the fringe width $(W)$.
$\therefore 8 W =0.12 \ cm\ ($by the data$)$
$\therefore W =\frac{0.12}{8} \ cm =0.015 \ cm =0.015 \times 10^{-2} m$
The distance $(D)$ between the slit and the eyepiece is equal to the sum of the distance between the slit and the biprism and the distance between the biprism and the eyepiece.
$\therefore D =20+80=100 \ cm =1 m\ ($by the data$)$
Also, $d _1=4.5\ mm$ and $d _2=2\ mm$
$\therefore$ The distance $(d)$ between the virtual images of the slit is
$d =\sqrt{d_1 d_2}= mm =3\ mm$
$=3 \times 10^{-3} m$
$\therefore$ The wavelength of light,
$d=\sqrt{d_1 d_2} =\sqrt{4.5 \times 2}\ mm =3\ mm$
$ =3 \times 10^{-3} m$
$\therefore$ The wavelength of light,
$\lambda =\frac{W d}{D}=\frac{0.015 \times 10^{-2} \times 3 \times 10^{-3}}{1}$
$ =4.5 \times 10^{-7} m$
$ =4.5 \times 10^{-7} \times 10^{10} \mathring A $
$ = 4 5 0 0 \mathring A$

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