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  • When a light of wavelength in vacuum travels through the same thickness

When a light of wavelength \mathrm{4000 \, \AA} in vacuum travels through the same thickness in diamond and water separately, the difference in the number of waves is 200 . Find the thickness, if refractive indices of diamond and water are \mathrm{\frac{5}{2}} and  \mathrm{\frac{4}{3}} respectively:

Option: 1

0.685 mm


Option: 2

0.0685 mm


Option: 3

68.5 mm


Option: 4

6.85 mm


Answers (1)

best_answer

Here, \mathrm{\mu_{\text {diamond }}=\frac{5}{2}, \mu_{\text {water }}=\frac{4}{3}}

\mathrm{ \lambda_{\text {vacuum }}=400 \AA=4000 \times 10^{-10} \mathrm{~m} }

Refractive index of diamond \mathrm{\mu_{\text {diamond }}=\frac{\text { Wavelength of light in vacuum }}{\text { Wavelength of light in diamond }}=\frac{\lambda_{\text {vacuum }}}{\lambda_{\text {diamond }}}}

\mathrm{ \lambda_{\text {diamond }}=\frac{4000 \AA}{(5 / 2)}=1600 \AA }
Refractive index of water \mathrm{\mu_{\text {water }}=\frac{\text { Wavelength of light in vacuum }}{\text { Wavelength of light in water }}=\frac{\lambda_{\text {vacuum }}}{\lambda_{\text {water }}}}

\mathrm{ \lambda_{\text {water }}=\frac{4000 \AA}{(4 / 3)}=3000 \AA }
Number of waves in thickness t of diamond, \mathrm{\mathrm{n}_{\text {diamon }}=\frac{\mathrm{t}}{\lambda_{\text {diamond }}}}

Number of waves in same thickness t of water, \mathrm{\mathrm{n}_{\text {water }}=\frac{\mathrm{t}}{\lambda_{\text {water }}}}

According to question, \mathrm{\mathrm{n}_{\text {diamond }}-\mathrm{n}_{\text {water }}=200}

\mathrm{ \begin{aligned} & \frac{\mathrm{t}}{\lambda_{\text {diamond }}}-\frac{\mathrm{t}}{\lambda_{\text {water }}}=200 \\\\ & \mathrm{t}\left(\frac{\lambda_{\text {water }}-\lambda_{\text {diamond }}}{\lambda_{\text {diamond }} \lambda_{\text {water }}}\right)=200 \\\\ & \mathrm{t}\left(\frac{3000 \AA-1600 \AA}{(1600 \AA)(3000 \AA)}\right)=200 \end{aligned} }
On solving, we get

\mathrm{ \mathrm{t}=6.85 \times 10^{-5} \mathrm{~m}=0.0685 \mathrm{~mm} }

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Gaurav

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