Question

Тонкая линза с оптической силой D = 10,0 дптр дает изображение предмета на экране, увеличенное в четыре раза. Определите расстояние от предмета до линзы.

Answer 1

1/d+1/f=D
f/d=4 f=4*D
1/d+1/4*d+D
5/4*d=D   d=5/4*D=5/4*10=0,125 м=12,5 см

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 of the lens (D) is 10.0 diopters and the image on the screen is magnified four times. We need to find the object distance (u).

To solve this problem, we can use the relationship between the optical power of the lens and the focal length:

D = 1/f

Therefore, the focal length (f) can be calculated as:

f = 1/D

Substituting the given value of D = 10.0 diopters, we find:

f = 1/10.0 = 0.1 meters

Now, we can use the lens formula to find the object distance (u). Since the image is magnified four times, the image distance (v) can be expressed as:

v = 4u

Substituting the values into the lens formula, we have:

1/f = 1/v - 1/u

1/0.1 = 1/(4u) - 1/u

Simplifying the equation:

10 = 1/4u - 1/u

10 = (1 - 4)/4u

10 = -3/4u

u = -3/40 meters

Therefore, the distance from the object to the lens is -3/40 meters.

Please note that the negative sign indicates that the object is located on the same side as the lens, which is consistent with the formation of a magnified image.