Question

Electron accelerate by potential difference of 250 V enter the electric field at angle of incidence of 50 degree and get refracted through an angle of 30 degree . Find the potential of second region. *
585.2 V
600 V
700 V
500 V​

Answer 1

Answer:

585.2 V.

Explanation:

given:- V1=250

V2=?

theta1=50

theta2=30

formula:-sin theta1/ sin theta2 =√V2/V1

sin50/sin30=√V2/250

V2=585.2V

Answer 2

Answer:

The potential of the second region is $433 V$. None of the options provided match this value, so there may be an error in the question or the answer choices.

Explanation:

We can use Snell's law to relate the angles of incidence and refraction to the refractive index of the second region:

$n1 sinθ1 = n2 sinθ2$

where n1 is the refractive index of the first region (which we can take to be air, and so is equal to 1), θ1 is the angle of incidence (50 degrees), n2 is the refractive index of the second region, and θ2 is the angle of refraction (30 degrees).

Rearranging this equation, we get:

$n2 = n1 sinθ1 / sinθ2\\n2 = sin50 / sin30\\n2 = 1.732$

We can then use the formula for the potential difference to find the potential of the second region:

$V = n2 V1$

where V1 is the potential of the first region $(250 V)$.

$V = 1.732 x 250\\V = 433 V$

Therefore, the potential of the second region is $433 V$. None of the options provided match this value, so there may be an error in the question or the answer choices.

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