Question

Ф И З И К А - Л И Н З А: При помощи линзы, фокусное расстояние которой 20 см, получено изображение предмета на экране, удалённом от линзы на 1 м. На каком расстоянии от линзы находится предмет? Каким оно будет?

Answer 1

Дано
F=20 см
f=100 см
d-?            формула тонкой собирающей линзы
1/F=1/d+1/f    1/20=1/d+1/100   d=25 см Изображение увеличено в 4 раза

Answer 2

This is a question about physics and lenses. According to the thin lens formula, the relationship between the object distance (p), the image distance (q), and the focal length (f) of a lens is given by:

$$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$$

In this case, the focal length of the lens is 20 cm, and the image distance is 1 m. We can plug these values into the formula and solve for the object distance:

$$\frac{1}{p}+\frac{1}{100}=\frac{1}{20}$$

$$\frac{1}{p}=\frac{1}{20}-\frac{1}{100}$$

$$\frac{1}{p}=\frac{4}{100}-\frac{1}{100}$$

$$\frac{1}{p}=\frac{3}{100}$$

$$p=\frac{100}{3}$$

$$p\approx 33.3 \text{ cm}$$

Therefore, the object distance is approximately 33.3 cm. To determine the characteristics of the image, we can use the magnification formula, which is given by:

$$m=-\frac{q}{p}$$

Plugging in the values of q and p, we get:

$$m=-\frac{100}{33.3}$$

$$m\approx -3$$

The magnification is negative, which means that the image is inverted. The absolute value of the magnification is 3, which means that the image is three times larger than the object. Thus, the image is inverted, enlarged, and real (since it can be projected on a screen).