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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.