All categories

3 years ago

When all else remains the same, what effect would decreasing the focal length have on a convex lens?

Physics

2 answers:

murzikaleks [220]3 years ago

8 0

Answer: The correct answer is "it would make the lens stronger".

Explanation:

The power of the lens is defined as the reciprocal of the focal length. If the focal length of the smaller then the power of the lens will more and vice versa.

The power of the lens is measured in Diopter.

More the power of the lens, the stronger the lens would be. The power of the lens has an ability to bend the light. Greater the power of the lens, more the refraction of the light will be.

If the focal length of the convex lens is decreased then the convex lens would make the lens stronger.

Therefore, the correct option is "it would make the lens stronger".

aniked [119]3 years ago

4 0

<h3>Answer;</h3>

It would make the lens stronger.

<h3>Explanation;</h3>

The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
Thus; decreasing the focal length of a convex lens makes the lens stronger.

You might be interested in

In a fit of rage about distance learning, you throw your computer out the window with an initial velocity of 3 m/s and it breaks

lesya692 [45]

Answer:

9 m

Explanation:

i did the test and got 100%

3 0

3 years ago

There are 2.2 pounds in a kilogram. if a boys mass is 40kg, what is his height in pounds?

sasho [114]

The simplest way to do this is to set up equivalent fractions, like this-

$\frac{1}{2.2}$ = $\frac{40}{x}$

Solve for x by using cross multiplication.

40*2.2= 88
1*x=88
x=88

Therefore, the boy weighs 88lbs.

3 0

3 years ago

Read 2 more answers

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th

jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

R = Vo² sin (2θ) / g

sin 2θ = g R / Vo²

sin 2θ = 9.8 75/35²

2θ = sin⁻¹ (0.6)

θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

X = Vox t

t2 = X2 / Vox = X2 / (Vo cosθ)

t2 = 37.5 / (35 cos 18.4)

t2 = 1.13 s

With this time we calculate the height at this point

Y = Voy t - ½ g t²

Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²

Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0

3 years ago

Read 2 more answers

The rectangular coordinates of a point are (5.00, y) and the polar coordinates of

Otrada [13]

Answer:

Explanation:

Polar coordinates formula

Summary. To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) : x = r × cos( θ ) y = r × sin( θ )

6 0

3 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos: