Suppose that intensity of a laser is (315/π) W/m². The rms electric field, in units of V/m associated with this source is close to the nearest integer is (ε₀ = 8.86 × 10⁻¹² C²N⁻¹m⁻²; c = 3 × 10⁸ ms⁻¹)
Suppose that intensity of a laser is (315/π) W/m². The rms electric field, in units of V/m associated with this source is close to the nearest integer is (ε₀ = 8.86 × 10⁻¹² C²N⁻¹m⁻²; c = 3 × 10⁸ ms⁻¹)
-
1 Answer
-
I=½ε? E? ²c ⇒ E? =√ (2I/ε? c)
Erms = E? /√2 = √ (I/ε? c) = √ (315/π / (8.86e-12 × 3e8) = 194
Similar Questions for you
According to definition of displacement current, we can write
B0 = 42.9 × 10-9
42.9 Ans
s? = -? + k? ∴ s? = (-? +k? )/√2
I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
From (i) and (ii)
P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9×10? ×2×15)/ (15×15×10? ×3×10? ) = 2×10? ³ V/m
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers