Physics Wave Optics
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New answer posted
2 months agoContributor-Level 10
Is? = 4I?
Is? = I?
∴ I_max/I_min = (√Is? + √Is? )² / (√Is? - √Is? )² = (2+1)²/ (2-1)² = 9/1
New answer posted
2 months agoContributor-Level 10
Factual (theory based)
Capacitor in parallel removes the AC ripple from the rectified output.
New answer posted
2 months agoContributor-Level 9
Fringe width, β = (λD) / d = (5890 * 10? ¹? * 0.5) / (0.5 * 10? ³) = 589 * 10? m
Distance between first and third bright fringe = 2β = 1178 * 10? m.
New answer posted
2 months agoContributor-Level 10
We know, P = √2Em and λ = h/√2Em
For 1st maxima, 2nd sinθ = λ
Put (1) and get the answer
E = 50.47eV.
B = µ? I (200)/2R? R? = 20 cm; R? = 1 cm
e = -dΦ/dt = π (R? )² (500) dB/dt
= π (R? )² (500) (µ? I/2R? ²) (200) (10t-2)
= π (1/10000) (500) (200) (µ? ) (8) (5)/ (2)
= 16π² x 10 x 5 x 10?
= 800π² x 10?
e = 0.7887mv
4/x = 0.8; x = 5.
New answer posted
2 months agoContributor-Level 10
We know I = 4I? cos² (φ/2) but φ = 2πx/λ
I = 4I? cos² (πx/λ)
(i) when x = λ, I = k
i.e. k = 4I? cos²π
k = 4I?
(ii) when x = λ/6
I' = kcos² (π (λ/6)/λ) = kcos² (π/6) = k (√3/2)²
I' = 3k/4
Correction from image: The provided image has an error in the calculation. I' = k ( (√3/2)² ) = 3k/4. The image shows 9k/12 which simplifies to 3k/4. I will use the value from the OCR which seems to be a typo.
I' = kcos² (π/6) = k (3/4)
I' = 9k/12
New answer posted
2 months agoContributor-Level 10
λ = 6 * 10? M
d = 6 x 10? M
I = I? [ (sin (β)²/β²] ; β = (πdsinθ)/λ
θ = π/2 ; β = πd/λ
= (π (6 * 10? )/ (6 x 10? ) = 100π
So at ∞ also minima will form total number of minima = 2 * 100 = 200
New answer posted
2 months agoContributor-Level 10
E = (I)(t)(A)(cos²θ)
⇒ (3.3) (2π / [31.4]) [3 * 10?] * 1/2
⇒ 0.99 * 10?
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