A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is
where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5
So we find
And given the distance of the screen from the slits,
The distance of the 5th bright fringe from the central bright fringe will be given by
B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:
To find the angle corresponding to the 8th dark fringe, we substitute n=8:
And the distance of the 8th dark fringe from the central bright fringe will be given by
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Answer: The potential difference between the plates = 0.4061V
Explanation:
Given that the
Electric field strength E = 155 N/C
Distance d = 0.00262 m
From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is
E = V/d
Substitute E and d into the above formula
155 = V/0.00262
Cross multiply
V = 155 × 0.00262
V = 0.4061 V
The potential difference between the plates is 0.4061 V