A light source is located 2.0 m below the surface of a...

A light source is located $2.0 \mathrm{m}$ below the surface of a swimming pool and $1.5 \mathrm{m}$ from one edge of the pool, as shown in Figure $18-26 .$ The pool is filled to the top with water. a. At what angle does the light reaching the edge of the pool leave the water? b. Does this cause the light viewed from this angle to appear deeper or shallower than it actually is? (FIGURE CANNOT COPY)

Transcript

00:01 Hi, here in this question we have given a source is placed at 2 meter below the swimming pool.

00:08 This is swimming pool contained water and we have to find out the angle of reflection with the water.

00:19 Now this source is placed from one edge 1 .5 meter away.

00:26 Now we have this angle, this is the triangle.

00:30 A s b from here we can find out the tangent theta tangent theta is basically perpendicular by base for this triangle perpendicular is this 1 .5 meter and the base is 2 meter so tangent theta is perpendicular by base so we can write 1 .5 by 2 so it will become 0 .75 from here theta will become 10 inverse 0 .75.

01:10 Now, using snell's law, we know, mu of water into sine theta is equal to mu of air into sine r.

01:35 So we have two medium water and air.

01:38 So it will become mu of water into sine theta and is equal to mu of air into sine r.

01:44 According to the snail's law now we know the refractive index of water is 1 .33 so we can put the value of muve water and for air it is 1 so it is 1 .33 and sine and theta is 10 inverse 0 .75 is equal to this is 1 into sine r we are using this 4 .75 we are using this 4 .7.

02:16 Because we have to find the r.

02:19 Now from here, so this is 1 .33 sine in 10 inverse 0 .75 is to sign r.

02:34 Now if we have to find out the value of r, we have to take sine as sine inverse 1 .33 .s.

02:44 Theta is 10 inverse 0 .75 is equal to r...

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