Question

Тонкая линза с оптической силой D = 7,5 дптр дает изображение предмета на экране, уменьшенное в пять раз. определить расстояние от предмета до линзы

Answer 1

D/f=5
f=d/5
1/F=1/d+1/f=1/d+5/d=6/d=D
d=6/D
d=6/7,5=0,8м
d-расстояние от предмета до линзы

Answer 2

Calculation of the Distance from the Object to the Lens

To determine the distance from the object to the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where: - f is the focal length of the lens - v is the image distance (distance from the lens to the image) - u is the object distance (distance from the lens to the object)

In this case, we are given that the optical power (D) of the lens is 7.5 diopters and the image formed on the screen is five times smaller than the object.

Let's assume the distance from the object to the lens is 'u' and the distance from the lens to the image is 'v'. Since the image is formed on the screen, the distance from the lens to the image is equal to the distance from the lens to the screen.

Given that the image is five times smaller, we can write:

v = 5u

Now, let's substitute the values into the lens formula:

1/7.5 = 1/(5u) - 1/u

Simplifying the equation:

1/7.5 = (1 - 5)/5u

1/7.5 = -4/5u

Cross-multiplying:

5u = -4 * 7.5

5u = -30

Dividing by 5:

u = -6

Since distance cannot be negative, we can ignore the negative sign. Therefore, the distance from the object to the lens is 6 units.

Please note that the units of distance are not specified in the question, so we cannot determine the exact unit of measurement.