Question

Найти оптическую силу и фокусное расстояние собирающей линзы,если действительное изображение предмета,помещенного на расстоянии 30 см от линзы,получается на расстоянии 60 см от нее.Найти увеличение!пожааалуйста,помогиитее!!

Answer 1

Calculation of Optical Power and Focal Length

To find the optical power and focal length of a converging lens, given that a real image is formed at a distance of 60 cm from the lens when the object is placed 30 cm away, 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 - u is the object distance from the lens

In this case, the object distance (u) is 30 cm and the image distance (v) is 60 cm.

Substituting the values into the lens formula:

1/f = 1/60 - 1/30

Simplifying the equation:

1/f = (2 - 1)/60

1/f = 1/60

Therefore, the focal length of the converging lens is 60 cm.

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

P = 1/f

Substituting the value of the focal length:

P = 1/60

Therefore, the optical power of the converging lens is 0.0167 diopters.

Calculation of Magnification

The magnification (m) of a lens is given by the formula:

m = -v/u

Where: - v is the image distance from the lens - u is the object distance from the lens

In this case, the object distance (u) is 30 cm and the image distance (v) is 60 cm.

Substituting the values into the magnification formula:

m = -60/30

Simplifying the equation:

m = -2

Therefore, the magnification of the lens is -2.

Please note that the negative sign indicates that the image formed is inverted.

I hope this helps! Let me know if you have any further questions.