Answer:
A

B

Explanation:
From the question we are told that
The wavelength is
The angle of first bright fringe is
The order of the fringe considered is n =1
Generally the condition for constructive interference is
=> 
=> 
Converting to cm

Generally the number of grating pre centimeter is mathematically represented as

=> 
=> 
Considering question B
From the question we are told that
The first wavelength is
The second wavelength is
The order of the fringe is 
The grating is 
Generally the slit width is mathematically represented as

=> 
=> 
Generally the condition for constructive interference for the first ray is mathematically represented as
=>
=> ![\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%20sin%5E%7B-1%7D%20%5B%5Cfrac%7B%202%20%2A%20%20%20650%20%2A10%5E%7B-9%7D%20%7D%7B%202%2A10%5E%7B-6%7D%7D%5D)
=> 
Generally the condition for constructive interference for the second ray is mathematically represented as
=>
=> ![\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20sin%5E%7B-1%7D%20%5B%5Cfrac%7B%202%20%2A%20%20%20420%20%2A10%5E%7B-9%7D%20%7D%7B%202%2A10%5E%7B-6%7D%7D%5D)
=> 
Generally the angular separation is mathematically represented as

=> 
=> 
I think you should hold a stretch for 10-30 seconds
Final speed = initial speed + (acceleration x time)
(final speed - initial speed) = acceleration x time
Time = (final speed - initial speed) / acceleration
In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up
First we will find the mass of the stone
As it is given that stone is spherical in shape so first we will find its volume



Now it is given that it's specific gravity is 10.8
So density of rock is

mass of the stone will be



now change in potential energy is given as

here
g = gravity on planet = 0.278 m/s^2
H = height lifted upwards = 15 cm


Now energy supplied by internal circuit of robot is given by

V = voltage supplied = 10 V
i = current = 1.83 mA
t = time = 12 s


Now efficiency is defined as the ratio of output work with given amount of energy used


so efficiency will be 23 %