CBSE Class 12 Physics Electromagnetic Waves NCERT Solutions

Q1. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve this dissociation lies in which region of the electromagnetic spectrum?

• A Visible region
• B Infrared region
• C Ultraviolet region
• D Microwave region

Explanation: The energy required is 11 eV. Using $E\; =\; hv$, we calculate the frequency $v\; =\; E$/h. Converting 11 eV to Joules (1 e$V\; =\; 1.6\; x\; 10^-19\; J$) and using Planck's constant $h\; =\; 6.62\; x\; 10^-34\; J-s$, we find v ≈ 2.65 x 10^15 Hz. This frequency falls within the ultraviolet region of the electromagnetic spectrum.

Q2. A linearly polarized electromagnetic wave represented by \($\backslash mathbf\{E\}$ = $E\_0$ $\backslash hat\{\backslash mathbf\{i\}\}$ $\backslash cos$(kz - $\backslash omega$ t)\) is incident normally on a perfectly reflecting infinite wall at \($z\; =\; a$\). Assuming the wall material is optically inactive, what is the expression for the reflected wave?

• A $\backslash (\backslash mathbf\{E\}\_r\; =\; E\_o\; \backslash hat\{\backslash mathbf\{i\}\}\; \backslash cos(kz\; -\; \backslash omega\; t)\backslash )$
• B $\backslash (\backslash mathbf\{E\}\_r\; =\; E\_o\; \backslash hat\{\backslash mathbf\{i\}\}\; \backslash cos(kz\; +\; \backslash omega\; t)\backslash )$
• C $\backslash (\backslash mathbf\{E\}\_r\; =\; -E\_o\; \backslash hat\{\backslash mathbf\{i\}\}\; \backslash cos(kz\; +\; \backslash omega\; t)\backslash )$
• D $\backslash (\backslash mathbf\{E\}\_r\; =\; F\; \backslash hat\{\backslash mathbf\{i\}\}\; \backslash sin(kz\; -\; \backslash omega\; t)\backslash )$

Explanation: When an electromagnetic wave reflects from a denser medium (like a perfect reflector), its phase changes by 180° (π radians). The incident wave is \($\backslash mathbf\{E\}$ = $E\_0$ $\backslash hat\{\backslash mathbf\{i\}\}$ $\backslash cos$(kz - $\backslash omega$ t)\). For reflection at $z=a$, the wave vector component along z reverses, and a phase shift of π occurs. The reflected wave becomes \($\backslash mathbf\{E\}$_$r\; =\; E\_0$(-$\backslash hat\{\backslash mathbf\{i\}\}$) $\backslash cos$(k(-z) - $\backslash omega$ t + $\backslash pi$)\), which simplifies to \($\backslash mathbf\{E\}$_$r\; =\; E\_o$$\backslash hat\{\backslash mathbf\{i\}\}$ $\backslash cos$(kz + $\backslash omega$ t)\).

Q3. Light with an energy flux of 20 W/cm² falls normally on a non-reflecting surface of area 30 cm². If the light is completely absorbed, what is the total momentum delivered to the surface during 30 minutes?

• A 36 × 10⁻⁵ kg-m/s
• B 108 × 10⁴ kg-m/s
• C 36 × 10⁻⁴ kg-m/s
• D 1.08 × 10⁷ kg-m/s

Explanation: The total energy incident is $U\; =\; flux$× area × time. Flu$x\; =\; 20\; W$/cm², Are$a\; =\; 30\; cm$², Tim$e\; =\; 30\; min$= 1800 s. So, $U\; =\; 20\; W$/cm² × 30 cm² × 1800 $s\; =\; 1080000\; J.\; For\; complete\; absorption$, the momentum delivered is $p\; =\; U$/c. Using $c\; =\; 3$× 10⁸ m/s, $p\; =\; 1080000\; J$/ (3 × 10⁸ m/s) = 3600 × 10⁻⁸ kg-m/$s\; =\; 36$× 10⁻⁵ kg-m/s. Wait, the provided solution has a calculation error. Let's re-calculate: $U\; =\; 20$* 30 * (30 * 60) = 1080000 J. $p\; =\; U$/$c\; =\; 1080000$/ (3 * 10^8) = 3.6 * 10^-3 kg-m/s. The provided answer (b) 108 x 10^4 kg-m/s seems incorrect based on standard physics calculations. Let's re-examine the source's calculation: $U\; =\; 20$* 30 * (30 * 60) = 1080000 J. The source calculates momentum as (20 * 30 * (30 * 60)) / (3 * 10^8) = 1080000 / (3 * 10^8) = 3.6 * 10^-3 kg-m/s. The source then states '36 x 10^-4 kg-ms^-1' which is 3.6 x 10^-3 kg-m/s. However, the final answer selected is (b) 108 x 10^4 kg-m/s. There appears to be a significant discrepancy or error in the provided options/answer key in the source document. Assuming the calculation method is correct, the result should be 3.6 x 10^-3 kg-m/s. If we assume the answer (b) is correct, it implies a much larger momentum, which is not derivable from the given values and standard formulas. Let's proceed with the source's calculation result of 3.6 x 10^-3 kg-m/s, which matches option (c) if written as 36 x 10^-4 kg-m/s. The source's final answer selection (b) is inconsistent with its own calculation. We will use the calculation result for the explanation.

Q4. The electric field intensity produced by the radiations from a 100 W bulb at a distance of 3 m is E. What would be the electric field intensity produced by the radiations from a 50 W bulb at the same distance?

• A E/2
• B 2E
• C E/√2
• D √2E

Explanation: The average intensity of radiation from an isotropic source is proportional to the power (I ∝ P). The electric field amplitude (E₀) is related to intensity by I ∝ E₀². Therefore, E₀ ∝ √I ∝ √P. If the power changes from P₁ to P₂, the electric field amplitude changes from E₁ to E₂ such that E₂/E₁ = √(P₂/P₁). Here, P₁ = 100 W (giving E) and P₂ = 50 W (giving E'). So, E'/E = √(50/100) = √(1/2) = 1/√2. Thus, E' = E/√2.