Question

Предмет находящийся на расстоянии 14см переместили на 6см от линзы, при этом увеличение стало равным единице. Определить оптическую силу этой линзы.

Answer 1

Если предмет оказался на расстоянии 20 см от линзы, то ответ 10 диоптрий. полный ответ во вложении.

Answer 2

Calculation of Optical Power of the Lens

To determine the optical power of the lens, we need to use the formula:

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

where the focal length is the distance at which parallel rays of light converge or diverge after passing through the lens.

In this case, we are given that an object located 14 cm away from the lens is moved to a distance of 6 cm from the lens. The magnification becomes equal to 1.

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

1/f = 1/v - 1/u

where: - f is the focal length of the lens, - v is the image distance from the lens, and - u is the object distance from the lens.

Given that the object distance u1 = 14 cm, object distance u2 = 6 cm, and magnification m = 1, we can solve for the focal length f.

Calculation Steps:

1. Calculate the image distance v1 using the magnification formula:

m = v1 / u1

Rearranging the formula, we get:

v1 = m * u1

Substituting the given values, we have:

v1 = 1 * 14 cm = 14 cm

2. Calculate the image distance v2 using the magnification formula:

m = v2 / u2

Rearranging the formula, we get:

v2 = m * u2

Substituting the given values, we have:

v2 = 1 * 6 cm = 6 cm

3. Substitute the values of v1, u1, v2, and u2 into the lens formula:

1/f = 1/v1 - 1/u1

1/f = 1/14 cm - 1/6 cm

Simplifying the equation, we get:

1/f = (6 - 14) / (14 * 6) = -8 / 84 = -1 / 10.5 cm^-1

Therefore, the optical power of the lens is -10.5 cm^-1.

Please note that the negative sign indicates that the lens is a diverging lens.