Question

MESSAGE HY INSTRUCTOR FULL SCREEN PRINTER VERSION 4BACK NEXT Chapter 34, Problem 078 GO More lenses. Object O stands on the central axis of a thin symmetric lens. For this situation (see the table below, all distances are in centimeters), find (a) the lens type, converging or diverging, (b) the focal distance f, (e) the image distance l. It also refers to whether (d) the image is real or virtual, and (e) on the same side of the lens as O or on the opposite side. (a) (ь) (e) (d) (e) Туре f Р R/V m 1/NI Side +13 0.40 NI (a) (b) Number Units (e) Number Units (d) (e) Click if you would like to Show Work for this question: Open Show Work

Answer 1

Hi,

Hope you are doing well.

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PART (a):

Given that the image is non-inverted (erect) and the magnification is positive and less than 1.

Therefore, the lens is Diverging lens (concave lens).

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PART (b):

Given that, magnification m = 0.4

Object distance, P = +13 cm.

Therefore, Image distance is given by:

m =

:1= -mP = -0.4 x 13 cm

:1= -5.2 cm

Focal length is given by:

1 7=ī 1 1 p= -5.2 cm 13 cm

if = -8.67 cm

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PART (c):

From PART (b): , Image distance is given by:

:1= -5.2 cm

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PART (d):

Since the image is formed by a diverging lens, the image is virtual.

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PART (e):

Image will be formed on the same side of the lens.

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