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Wave Optics: Diffraction and Interference - Class XII

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In a Fresnel biprism experiment, the two positions of lens give separation between the slits as $16 $cm and$9 $cm, respectively. What is the actual distance of separation?

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A
$12 cm$
💡 Explanation:

A Fresnel Biprism is the variation on the Young's Slits experiment. The Fresnel biprism has two thin prisms which are joint at their bases to form an isosceles triangle. A single wavefront impinges on both prisms.

Separations between the slits,

$d _{1}= 16 cm$

and $d _{2}= 9 cm.$

Actual distance of separation (d)  can be computed with the formula as given:

$d = \sqrt {d _{1} \times d _{2}}\\$

$d = \sqrt {16 \times 9}\\$

$d = \sqrt {144} \\$

$d = 12 cm$

Thus actual distance will be $12 cm$.

Option B is correct.

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