Question

Высота изображения предмета 40см ,расстояние от экрана ,на котором получено изображение ,до собирающей линзы 5см .Чему равна оптическая сила линзы,если высота предмета 18см?

Answer 1

Дано:
H=40 см
f=5 см
h = 18 см
____________
D - ?

Увеличение линзы:
Г = H/h = 40/18 = 20/9 

Но увеличение линзы можно найти и по формуле
Г = f/d
Тогда
d = f/Г = 5 / (20/9) = 5*9/20 = 9/4 = 2,25 см

Фокусное расстояние линзы:
F = f*d / (f+d) = 5*2,25 / (5+2,25) = 1,55 см   или 0,0155 м

Оптическая сила:
D = 1/F = 1/0,0155 ≈  65 дптр

(Что-то много... Проверьте, пожалуйста, условие! Может быть f=50 см ??? )

Answer 2

Calculation of Optical Power of the Lens

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

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

In this case, we are given the height of the object (h1) and the distance from the screen to the lens (d1). We need to find the optical power of the lens when the height of the object is 18 cm.

Let's calculate the focal length (f) of the lens using the given information:

1. The height of the object (h1) = 40 cm 2. The distance from the screen to the lens (d1) = 5 cm

Using the formula for magnification:

Magnification (M) = h2 / h1 = -d2 / d1

where: - h2 is the height of the image - d2 is the distance from the lens to the image

We can rearrange the formula to solve for d2:

d2 = -M * d1

Given that the height of the object (h1) is 40 cm and the distance from the screen to the lens (d1) is 5 cm, we can calculate the magnification (M) using the formula:

M = h2 / h1

Substituting the values, we get:

M = 18 cm / 40 cm = 0.45

Now, we can calculate the distance from the lens to the image (d2):

d2 = -0.45 * 5 cm = -2.25 cm

Since the image is formed on the same side as the object, the distance (d2) is negative.

Now, we can calculate the focal length (f) of the lens using the lens formula:

1 / f = 1 / d1 + 1 / d2

Substituting the values, we get:

1 / f = 1 / 5 cm + 1 / (-2.25 cm)

Simplifying the equation, we find:

1 / f = 0.2 cm^-1 - 0.44 cm^-1

1 / f = -0.24 cm^-1

Taking the reciprocal of both sides, we get:

f = -4.17 cm

The negative sign indicates that the lens is a diverging lens.

Therefore, the optical power of the lens is:

Optical Power (P) = 1 / f = 1 / (-4.17 cm) = -0.24 cm^-1

So, the optical power of the lens is approximately -0.24 cm^-1.

Please note that the calculations are based on the given information and assumptions.