VA and I will not do this if they don't want it and you are a little
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Answer:
Magnitude of induced emf is 0.00635 V
Explanation:
Radius of circular loop r = 45 mm = 0.045 m
Area of circular loop 
Magnetic field is increases from 250 mT to 350 mT
Therefore change in magnetic field 
Emf induced is given by
Magnitude of induced emf is equal to 0.00635 V
