Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that the speed of light is given as


now the frequency of the light is given as

so we have


Part b)
Position of Nth maximum intensity on the screen is given as

so here we know for 3rd order maximum intensity

n = 3
L = 1.4 m


Part c)
angle of third order maximum is given as



The answer is D. Because the boy jump 3 high
<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>