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katrin [286]
3 years ago
11

How do your results from ray tracing compare to your results from using the thin-lens equation?

Physics
1 answer:
EastWind [94]3 years ago
7 0

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} where;

f is the focal length  of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm

Hence the focal length of the convex lens is 20cm

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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine
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Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

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