Question

Тонкая линза дает изображение предмета на экране, Уменьшенное в четыре раза. Расстояние от предмета до линзы 1 метр. Определите оптическую силу линзы.

Answer 1

Увелиение равно расстояние от линзы до изображения делить на расстояние от линзы до предмета и по условию равно 1/4
тогда расстояние от линзы до изображения равно 25 см и оптическая сила равна 1/1 + 1/0,25 и равна 5 диоптрий

Answer 2

Calculation of Optical Power of the Lens

To determine the optical power of the lens, we can use the thin lens equation, which relates the object distance (denoted as "u"), the image distance (denoted as "v"), and the focal length of the lens (denoted as "f"). The equation is as follows:

1/f = 1/v - 1/u In this case, the object distance (u) is given as 1 meter, and the image distance (v) is given as 1/4 of the original size. We need to find the focal length (f) of the lens.

Let's substitute the given values into the thin lens equation and solve for the focal length:

1/f = 1/(1/4) - 1/1

Simplifying the equation:

1/f = 4 - 1

1/f = 3

Therefore, the focal length of the lens is 1/3 meter.

Calculation of Optical Power

The optical power of a lens is defined as the reciprocal of its focal length and is measured in diopters (D). The formula to calculate the optical power is:

Optical Power (P) = 1/f Substituting the value of the focal length (f) into the formula:

P = 1/(1/3)

Simplifying the equation:

P = 3/1

Therefore, the optical power of the lens is 3 diopters (D).

Please note that the calculation assumes a thin lens and uses the thin lens equation to determine the optical power.