Lets us consider an example:
Suppose a 10 ohm bulb is connected across the terminals of a 10 V
battery having 2ohm internal resistance.
Then total reistance in series we know, R1 + R2
Thus, R net = 10+ 2 = 12 ohm
The, current across circuit = 10/12= 0.833 A
Now, Power is given by 
Thus, power dissipated across internal resistance, P = 
And, total power dissipated =
Thsu, percentage of pwer not avaible = 1.37/8.2 = 16.70%
Answer:
5 fringes option C
Explanation:
Given:
- The wavelength of blue light λ = 450 nm
- The split spacing d = 0.001 mm
Find:
How many bright fringes will be seen?
Solution:
- The relationship between the wavelength of the incident light, grating and number of bright fringes seen on a screen is derived by Young's experiment as follows:
sin(Q) = n* λ / d
Where, n is the order of bright fringe. n = 0, 1, 2, 3, ....
- We need to compute the maximum number of fringes that can be observed with the given condition and setup. Hence we will maximize our expression above by approximating sin(Q).
sin(Q_max) = 1
Q_max = 90 degree
- Hence, we have:
n = d / λ
- plug values in n = 0.001 *10^-3 / 450*10^-9
n = 2.222
- Since n order number can only be an integer we will round down our number to n = 2.
- Hence, we will see a pair of bright fringes on each side of central order fringe.
- Total number of fringes = 2*2 + 1 = 5 fringes is total ... Hence, option C
