Question

Угол между падающим лучом и поверхность жидкости 60 градусов , показатель приломления жидкости 1,5. Чему равен синус угла преломления луча в этой жидкости ?

Answer 1

Угол падения
a=90-60=30
Закон преломления
sina/sinb= n
sin30/sinb=1,5
sinb=sin30/1,5=0,33
Ответ sinb=0,33

Answer 2

Calculation of the sine of the angle of refraction in the given liquid

To calculate the sine of the angle of refraction in the given liquid, we can use Snell's law, which relates the angles and refractive indices of the incident and refracted rays.

According to Snell's law, the relationship between the angles of incidence and refraction, denoted by α and β respectively, and the refractive indices of the two media, denoted by n₁ and n₂, is given by:

n₁ sin(α) = n₂ sin(β) (1)

In this case, the angle of incidence α is 60 degrees and the refractive index of the liquid is 1.5.

Substituting the given values into equation (1), we can solve for the sine of the angle of refraction β:

1 sin(60) = 1.5 sin(β)

Simplifying the equation:

sin(β) = (1 sin(60)) / 1.5

Using a calculator, we can evaluate the expression:

sin(β) ≈ 0.5774

Therefore, the sine of the angle of refraction in this liquid is approximately 0.5774.

Conclusion

The sine of the angle of refraction in the given liquid is approximately 0.5774.