### Physics help, two optics questions

The focal length of diverging lens is negative. If f= -20cm fro a particular diverging lens, where will the image be formed of an object located 50cm to the left of the lens on the optical axis? What is the magnification of the image?

The equation connecting s, p, and f for a simple lens can be employed for spherical mirrors, too. A concave mirror with a focal length of 8cm forms an image of a small object placed 10cm in front of the mirror. Where will this image be located?

Answers

a) The equation of the lens is

1/x1 +1/x2 = 1/f

where x1 is the position of the object, positive to the left of the lens

and x2 is the position if the image positive to the right positive to the lens

and f is positive for a converging lens and negative for a diverging lens

1/x2 =1/f-1/x1

1/x2 =-1/20 -1/50 =-70/100

x2 =-100/70 =-1.429 cm, therefore the image is to the left of the lens

The magnification is simply

y2/y1 = -x2/x1 = 1.429/50 =0.02958, the image is smaller and noninverted

b) the equation of a concave mirror is

1/p1 +1/p2 =1/f,

where p1 and p2 are the positions of the object, respecively image (positive if both to the left of the mirror) and f is positive for a concave mirror

1/p2 =1/f -1/p1 =1/8-1/10 =2/80

p2 =40 cm , to the left of the mirror, real image

the magnification is simply Y2/Y1 =-p2/P1 =-40/10 = -4, the image is bigger and inverted

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