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hichkok12 [17]
3 years ago
5

A ray of light travels across a liquid-to-glass interface. if the indices of refraction for the liquid and glass are, respective

ly, 1.75 and 1.52, what is the critical angle at this interface?
Physics
1 answer:
prisoha [69]3 years ago
7 0
When it comes to optics, Snell's law is the basic formula to be used. If you notice, when light hits the water, the light does not travel in the same direction. After, it hits the water, it changes in angle. Light becomes refracted. This is observed when your hands tend to become bigger if you place it underwater. The formula for Snell's Law is

n₁ sin θ₁ = n₂sin θ₂, where n is the index of refraction. This depends on the type of medium. For example, for air, n=1. The parameters θ₁ is the angle of incidence, and θ₂ is the angle of refraction. Critical angle is the incident angle needed so that the refract angle is 90°. So, modifying the equation:

n₁ sin θcrit = n₂sin 90°, since sin 90°=1,
sin θcrit = n₂/n₁
θcrit = sin ⁻¹ (n₂/n₁)

Since liquid comes first before glass, n₁=1.75 and n₂=1.52. Substituting,
θcrit = sin ⁻¹ (1.52/1.75)
θcrit = 60.29°

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When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t
Vaselesa [24]

Answer:

toward the normal

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.  

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

6 0
4 years ago
My Exams are coming.so, please tell me some ways to score good marks?​
Citrus2011 [14]

Instead of asking this question go and study

5 0
2 years ago
Read 2 more answers
A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog
DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

a)

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

b)

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

c)

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0
3 years ago
A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ
Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0
3 years ago
The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving
nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

F_f = \mu_s (mg)

where

\mu_s = 0.5645 is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N

So, the applied force must be greater than this value.

3 0
3 years ago
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