Question

1.) An object is placed in front of a diverging lens with a focal length of 17.7 cm. For each object distance, find the image distance and the magnification. Describe each image.

(a) 35.4 cm

location	_____cm
magnification	_____
nature	real virtual upright inverted

(b) 17.7 cm

location	_____  cm
magnification	_____
nature	real virtual upright inverted

(c) 8.85 cm

location	_____ cm
magnification	_____
nature	real virtual upright inverted

2.) An object is placed in front of a converging lens with a focal length of 15.0 cm. For each object distance, find the image distance and the magnification. Describe each image.

(a) 30 cm

location	____ cm
magnification	____
nature	real virtual upright inverted

(b) 7.5 cm

location	____cm
magnification	____
nature	real virtual upright inverted

Answer 1

u = object distance
v = image distance
f = focal length
m = magnification

1 / u + 1 / v = 1 / f
1 / 17 + 1 / v = 1 / (34)
v =x cm.
Image distance x cm.

(b)
m = v / u
= x / 17
= y.
Magnification y.

(c)
Virtual.

(d)
Upright.

(e)
Reduced.

Answer 2

here

focal length=17.7cm(f)

object distance=35.4cm(i)

image distance=TO FIND(I)

USING FORMULA :1/V+1/I=1/F

1>location=35.4cm

b> magnification=i/u=35.4/35.4=1

c>real

use above formulas to calculate all the answers

i provide u the natures of all the images

2>virtual

3>upright

4>reduced

Answer 3

for a diverging lens we have

-1/f = 1/u + 1/v

where f is focal length of lens,u is object distance and v is image distance

or 1/v = -1/f - 1/u

the magnification is

m = (v/u)