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Light of Wavelength 560 Nm Goes Through a Pinhole of Diameter 0.20 Mm and Falls on a Wall at a Distance of 2.00 M. What Will Be the Radius of the Central Bright Spot Formed on the Wall? - Physics

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प्रश्न

Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and  falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?

योग

उत्तर

Given:-

Wavelength of the light used,

\[\lambda = 560 nm = 560 \times {10}^{- 9} m\]

Diameter of the pinhole, d = 0.20 mm = 2 × 10−4 m

Distance of the wall, D = 2m

We know that the radius of the central bright spot is given by

\[R = 1 . 22\frac{\lambda D}{d}\]

\[       = 1 . 22 \times \frac{560 \times {10}^{- 9} \times 2}{2 \times {10}^{- 4}}\]

\[       = 6 . 832 \times  {10}^{- 3} m\text{ or }= 0 . 683  cm\]

Hence, the diameter 2R of the central bright spot on the wall is 1.37 cm.

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Width of Central Maximum
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अध्याय 17: Light Waves - Exercise [पृष्ठ ३८३]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 17 Light Waves
Exercise | Q 40 | पृष्ठ ३८३
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