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  • A beam of light consisting of two wavelengths 7000 Å and 5500 Å is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.

A beam of light consisting of two wavelengths 7000 Å and 5500 Å is used to obtain interference pattern in
Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the place of
slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both
the wavelengths coincide, is n × 10-5 m. The value of n is ______.

Option: 1

462


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let n1 maxima of 7000 Å coincides with n2 maxima of 5500 Å
therefore  \mathrm{n}_1 \beta_1=\mathrm{n}_2 \beta_2

\text { or } \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_2}{\lambda_1}=\frac{5500}{7000}=\frac{11}{14}

therefore 11th maxima of 7000 Å will coincide with 14th maximum of 5500 Å
To find the least distance of this
y = n1\beta1

$$ \begin{aligned} & \text { or } \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}} \\ & =\frac{11 \times 7 \times 5}{2.5} \times 10^{-5} \mathrm{~m} \\ & \text { or } \mathrm{y}=462 \times 10^{-5} \mathrm{~m} \\ & \text { therefore } \mathrm{n}=462 \end{aligned}

Posted by

avinash.dongre

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