The central fringe of interference pattern produced by light of wavelength 6000 A is found to shift to the position of 4th bright fringe- after a glass plate of -1-5 is introduced- The thickness of the glass plate is-

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Standard XII

Physics

Matter Waves

The central f...

Question

The central fringe of interference pattern produced by light of wavelength $6000 Å$ is found to shift to the position of $4^{th}$ bright fringe, after a glass plate of $μ = 1.5$ is introduced. The thickness of the glass plate is:

4.8 μ m

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8.23 μ m

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14.98 μ m

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3.78 μ m

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Solution

The correct option is A $4.8 μ m$
We know,
Shift at $n^{th}$ bright fringe is
$(μ - 1) t = n λ$
Given, $μ = 1.5$
$n = 4$
$λ = 6000 Å$
So, $(1.5 - 1) t = 4 (6000 \times 10^{- 10})$
$t = 4.8 \times 10^{- 6} m$
$\Rightarrow t = 4.8 μ m$
So, option (a) is correct.

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The central fringe of interference pattern produced by light of wavelength 6000 Å is found to shift to the position of 4th bright fringe, after a glass plate of μ=1.5 is introduced. The thickness of the glass plate is:

The correct option is A 4.8 μmWe know, Shift at nth bright fringe is (μ−1)t=nλ Given, μ=1.5 n=4 λ=6000 Å So, (1.5−1)t=4(6000×10−10) t=4.8×10−6 m ⇒t=4.8 μm So, option (a) is correct.

The central fringe of interference pattern produced by light of wavelength $6000 Å$ is found to shift to the position of $4^{th}$ bright fringe, after a glass plate of $μ = 1.5$ is introduced. The thickness of the glass plate is:

The correct option is A $4.8 μ m$
We know,
Shift at $n^{th}$ bright fringe is
$(μ - 1) t = n λ$
Given, $μ = 1.5$
$n = 4$
$λ = 6000 Å$
So, $(1.5 - 1) t = 4 (6000 \times 10^{- 10})$
$t = 4.8 \times 10^{- 6} m$
$\Rightarrow t = 4.8 μ m$
So, option (a) is correct.