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AleksAgata [21]

3 years ago

15

If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o

f the lens, what is the height of the image?
A. 14 millimeters
B. 6 millimeters
C. 22 millimeters
D. 8 millimeters

2 answers:

xz_007 [3.2K]3 years ago

6 0

As per the question the height of object $[h_{0} ]$ 18 mm

The object distance [u] is given as 12 mm

The image distance[v] is given as 4 mm

The lens taken here is a diverging nature.hence the lens is a concave lens.

The image formed in a concave lens is always virtual,erect and diminished.

As per the question the image is formed in front of the lens.

we have to calculate the image height $[h_{i} ]$ which is calculated as follows-

putting the sign convention on the above data we get

u= -12 mm v=-- -4 mm [ - sign is due to the fact that the measurement is opposite to the direction of light

$h_{0} =18 mm$ [the height is positive as it is above the principal axis]

As per the magnification formula we know that

Magnification $m= \frac{h_{i} }{h_{o} } = \frac{v}{u}$

⇒ $\frac{h_{i} }{h_{o} } =\frac{v}{u}$

⇒ $\frac{h_{i} }{18} =\frac{-4}{-12}$

⇒ $h_{i} =\frac{18*[-4]}{[-12]}$

= 6 mm [ans]

Hence B is right.

andrezito [222]3 years ago

4 0

The answer is

18 / x = 12 / 4
12x = 72
x = 6mm

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