An electromagnetic wave traveling in a

A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui

GalinKa [24]

Answer:

$n_l = 1.97$

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

$\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

focal length of lens in liquid is

$\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}$

rearrange fro $n_l$

$n_l = \frac{n_g f_l}{f_l+f(n_g-1)}$

$n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}$

$n_l = 1.97$

7 0

3 years ago

What is the energy in an atoms nucleus

olya-2409 [2.1K]

Answer:

the atomic energy is energy carried by atoms.

Explanation:

this is either Nuclear Binding energy...( the energy required to split the nucleus of the atom.) and the Nuclear Potential energy (the total energy of the particles inside the atomic nucleus.)

4 0

3 years ago

Why is it difficult to see colors outside at night?

Stella [2.4K]

I choose D. but I'm not sure if it's right yet. Let me know if it was :)

7 0

3 years ago

Which type of wave is classified as electromagnetic?

Nimfa-mama [501]

Answer: B) Light

Explanation: light waves are Electromagnetic waves. Visible light is one of the many types of electromagnetic waves. The others are mechanical waves.

8 0

3 years ago

Read 2 more answers

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

Given data:

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

To find:

Final velocity: ?

Formula:

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

Solution:

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago