Light travels from air (n = 1) into a tank of water (n = 1.33) and continues on into a layer of glass (n = 1.5). the light strikes the air-water interface at θi = 45°. what angle does the beam make to the normal line inside the glass (i. e., θr in the glass)? a) 20° b) 28° c) 36° d) 44°

option B is correct

beam make to the normal line inside the glass angle is  28°

Explanation:

Given data

air (n = 1)

water (n = 1.33)

glass (n = 1.5)

air-water interface at θi = 45°

to find out

angle does the beam make to the normal line inside the glass i.e θr

solution

we know here in air water interface

n1 = 1 and angle 45° and n2 = 1.33

n1×Sin(θ)i = n2×Sin(θ)r      ...............1

so now put all value in equation 1

1×Sin45 = 1.33× sin(θ)r  

(θ)r  = 32.12°

so this is here (θ)r  refraction angle in the water

and

we know this angle act as incident angle for water glass

so that

n2×Sin(θ)r  = n3×Sin(θ)g    ....................2

here n3 = 1.5  and n2 = 1.33 and (θ)r  = 32.12°

so

1.33 × Sin 32.12 = 1.5×Sin(θ)g

(θ)g = 28 degree

so option B is correct

beam make to the normal line inside the glass angle is  28°

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j