Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>x-rays</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>micro waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>radio waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>visible</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>visible</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>micro waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>radio waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&lt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>x-rays</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>visible</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>x-rays</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>radio waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>micro waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>radio waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>micro waves</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>visible</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> <mo>&gt;</mo> <msub> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mtext>x-rays</mtext> <mtext>&nbsp;</mtext> </mrow> </mrow> </msub> </math> </span></p>
3 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 4
Detailed Solution:

Kindly go through the solution 

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The angle between the plane of vibration and plane of polarization is 90°.

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
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Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
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3d = 0.6mm

D = 80cm

= 800mm

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BP – Andhra Pradesh = Dx

d y D = ( 2 n + 1 ) λ 2   [for Dark fringe at P]

n = 0, for first dark fringe

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λ = 2 d D y

= 2 d D × d 2 [ y = d 2 , G i v e n  first dark fringe is observed on the screen directly opposite to one of the slits]

λ = 2 × 0 . 6 × 0 . 6 8 0 0 × 2

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Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025

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