Question

Question 1:

A near-sighted person might correct his vision by wearing diverging lenses with focal length f = -50cm . When wearing his glasses, he looks not at actual objects but at the virtual images of those objects formed by his glasses. Suppose he looks at a 12cm -long pencil held vertically 2.0m from his glasses. Use ray tracing to determine the location of the image. Answer is in meters.

Question 2:

A 1.0-cm-tall object is 60 cm in front of a diverging lens that has a -30 cm focal length.

a) Calculate the image position in cm.

b) Calculate the image height in cm.

Show work please! Thank you!!

Answer 1

ho= 12cm, do = 200 cm, f = -50 cm

A) 1/f =1/do + 1/di
1/-50 = 1/200 + 1/di, :

di=-40cm

B) m = hi / ho = - di / do
m = -di / do = -0.2
hi = 12cm * .2 = 2.4 cm

2)

a) you'll need the mirror equation;
1/f = 1/u + 1/v -------------(1)
and for your question, u=object distance=60cm, f=focal lenght= -30cm (the value is negative because the mirror is convex, it will be positive for concave mirrors)

if you substitute those values into equation (1), you should get the image distance v as -20cm;
(the negative sign shows that the image is located behind the mirror surface, that also answers the question asked next. If the value was positive, then it shows the image is in front of the mirror surface, and a value of zero will have shown that the image is at the mirror surface)

(b) here we need the magnification formula;
image height/object height = image distance/object distance

if you put the values of object height=1.0cm, image distance= -20cm, object distance=60cm, then you'll get the image height as 0.33cm
(the negative sign still shows the image is virtual, and so upright; ALL IMAGES FORMED BY CONVEX MIRRORS ARE VIRTUAL AND UPRIGHT)