Question

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

Answer 1

Γ=5                            оптическая сила: D=1/d+1/f;  (1)
d=10 см=0,1 м           увеличение: Γ=f/d  ⇒  f=Γd; подставляем в (1),
_____________           D=1/d+1/Γd=1/0,1+1/5*0,1=10+2=12 дптр;
D-?                                 Ответ: D=12 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 of the object formed by the lens is magnified 5 times and the distance from the object to the lens is 10 cm.

Using the given information, we can calculate the focal length of the lens and then determine its optical power.

Calculation Steps:

1. Given: - Magnification (M) = 5 - Distance from object to lens (d) = 10 cm

2. Calculate the focal length (f) of the lens using the magnification formula: - M = -d / f - Substitute the given values: 5 = -10 / f - Rearrange the equation to solve for f: f = -10 / 5 = -2 cm

3. Calculate the optical power (P) of the lens using the formula: - P = 1 / f - Substitute the value of f: P = 1 / (-2 cm) = -0.5 diopters (D)

Therefore, the optical power of the lens is -0.5 diopters (D).

Please note that the negative sign indicates that the lens is a diverging lens, which causes light rays to spread out.