Question

Тонкая линза дает изображение на экране увеличенное в 5 раз.Расстояние от предмета до линзы d-10см, определите оптическую силу линзы

Answer 1

По формуле F=1/d, подставим числовые значения F=1/10; F=0,1 дптр

Answer 2

Calculation of Optical Power of a Lens

To determine the optical power of a lens, we can use the formula:

Optical Power (P) = 1 / Focal Length (f)

In this case, we are given that the image formed by the lens is magnified 5 times and the distance from the object to the lens is 10 cm (d = 10 cm).

To find the focal length of the lens, we can use the magnification formula:

Magnification (M) = -Image Distance (di) / Object Distance (do)

Given that the magnification is 5 times (M = 5), we can rearrange the formula to solve for the image distance:

di = -M * do

Substituting the given values, we have:

di = -5 * 10 cm = -50 cm

Since the image distance is negative, it means that the image is formed on the same side as the object, which indicates a diverging lens.

Now, we can use the lens formula to find the focal length:

1 / f = 1 / do + 1 / di

Substituting the values, we have:

1 / f = 1 / 10 cm + 1 / (-50 cm)

Simplifying the equation, we get:

1 / f = 1 / 10 cm - 1 / 50 cm

1 / f = (5 - 1) / 50 cm

1 / f = 4 / 50 cm

1 / f = 0.08 cm^(-1)

Therefore, the optical power of the lens is:

P = 1 / f = 1 / 0.08 cm^(-1) = 12.5 cm^(-1)

So, the optical power of the lens is 12.5 cm^(-1) or 12.5 diopters.