Question

A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm, determine the image distance and the image size.

Answer 1

Answer:

Image distance is - 9 cm.

Image size is 1 cm

Explanation:

u = - 35.5 cm

f = - 12 cm

ho = 4 cm

Lens formulae : 1/v - 1/u = 1/f

1/v = - 1/12 - 1 /36

     = - 4/36

v = - 9 cm

hi /ho = v /u         (magnification)

hi / 4 = - 9 / - 36

hi = 1/4 x 4 = 1 cm

Answer 2

Answer:

Image distance ($d_{i}$) = - 9.08 cm

Image height ($h_{i}$) = 1.02 cm

Given: (object distance) $d_{o} = 35.5 cm$, (object height) $h_{o} = 4.00cm$, (focal length) $f = -12.2$

Required: image distance ($d_{i}$) and image size ($h_{i}$)

Equations:

Eqn 01: $\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}}$ (mirror equation)

Eqn 02: $\frac{-d_{i}}{d_{o}} = \frac{h_{i}}{h_{o}}$ (magnification equation)

Solution:

Solving for the image distance:

1. $\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}}$
2. $\frac{1}{-12.2} = \frac{1}{35.5} + \frac{1}{d_{i}}$
3. $\frac{1}{-12.2} - \frac{1}{35.5} = \frac{1}{d_{i}}$
4. $-0.110136227 = \frac{1}{d_{i}}$
5. $d_{i} = {1}/{ -0.110136227}$
6. $d_{i} = -9.07966457cm$

Solving for the image height / size:

1. $\frac{-d_{i}}{d_{o}} = \frac{h_{i}}{h_{o}}$
2. $\frac{9.07966457}{35.5} = \frac{h_{i}}{4.00}$
3. $\frac{9.07966457*4.00}{35.5} = h_{i}$
4. $h_{i} = 1.023060797cm$

there are 3 significant figures (35.5 is 3, 4.00 is 3,  -12.2 is 3),  therefore,

Answer:

Image distance ($d_{i}$) = - 9.08 cm

Image height ($h_{i}$) = 1.02 cm