Question

in an experiment with a rectangular glass sla, a student observed that a ray of light incident at an angle of 55 with normal onone face of the slab, afterrefraction,strikes the opposite face of the glass slab before emerging out into air making an angle of 40 with the normal. what value would you assign to the angle of refraction and angle of emergence?

Answer 1

Answer:

thank you

Explanation:

i hope it helps u

Answer 2

Answer:

The angle of refraction r1 = $40^{0}$

The angle of emergence θ = $55^{0}$

Explanation:

From the above question,

They have given :

                   i = $55^{0}$

                   r2 = $40^{0}$

According to Snell's regulation of refraction, the ratio of the sine of the perspective of incidence to the sine of the perspective of refraction is steady for a given pair of media. Mathematically,

               sin(i) / sin(r) = constant

where i is the attitude of incidence.

Therefore, we can write:

               sin(55) / sin(r) = sin(40) / sin(e)

We can rearrange this equation to get:

               sin(r) = sin(55) * sin(e) / sin(40)

Using the cost of sin(55) = 0.819 and sin(40) = 0.643, we can calculate sin(r) as:

               sin(r) = 0.819 * sin(e) / 0.643

Solving for sin(e), we get:

               sin(e) = sin(r) * 0.643 / 0.819

Using the cost of sin(r), we can calculate sin(e) as:

               sin(e) = sin(55) * 0.643 / 0.819

               sin(e) = 0.627

Now, we can use Snell's regulation once more to locate the perspective of refraction 'r'. According to Snell's law:

sin(i) / sin(r) = n

where n is the refractive index of the glass slab.

We can expect that the refractive index of the glass slab is 1.5. Therefore, we can write:

               sin(55) / sin(r) = 1.5

Solving for sin(r), we get:

               sin(r) = sin(55) / 1.5 = 0.546

The angle of refraction r1 = $40^{0}$

The angle of emergence θ = $55^{0}$

For more such related questions : https://brainly.in/question/32667309

#SPJ3