Question

Расстояние от линзы с оптической силой 1.5 дптр до изображения предмета в два раза меньше,чем расстояние от предмета до линзы.Определите эти расстояния и линейное увеличение.

Answer 1

Дано f=2*d       D=1,5 дптр    d- ?   f- ?    Г - ?

Г=f/d=2*d/d=2

1/d+1/2*d=D
3/2*d=D
d=3/2*1.5=1 м
f=2 м

Answer 2

Calculation of Distances and Linear Magnification

To determine the distances and linear magnification in this scenario, we can use the lens formula and magnification formula.

Let's denote the distance from the lens to the object as u, the distance from the lens to the image as v, and the focal length of the lens as f.

According to the given information, the distance from the lens to the image is half the distance from the object to the lens. Mathematically, this can be expressed as:

v = (1/2)u

Using the lens formula, we have:

1/f = 1/v - 1/u

Substituting the value of v in terms of u:

1/f = 1/((1/2)u) - 1/u

Simplifying the equation:

1/f = 2/u - 1/u

Combining the terms on the right side:

1/f = 1/u

Therefore, we can conclude that:

u = f

Now, let's calculate the linear magnification (m). The linear magnification is given by the formula:

m = -v/u

Substituting the value of v in terms of u:

m = -((1/2)u)/u

Simplifying the equation:

m = -1/2

Therefore, the linear magnification in this scenario is -1/2.

To summarize: - The distance from the object to the lens (u) is equal to the focal length of the lens (f). - The distance from the lens to the image (v) is half the distance from the object to the lens. - The linear magnification (m) is -1/2.

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