Question

Rajashree wants to get an inverted image of height 5 cm of an object kept at
a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At
what distance from the mirror should she place the screen ? What will be the type of the
image, and what is the height of the object?​

Answer 1

Answer:

1/f=1/u+1/v

1/10-1/30=1/v

3-1/30=1/v

2/30=1/v

v=15cm

hi/ho=-v/h

1/ho=-v/u×hi

1/ho=-15/30×5

ho=-10cm

nature of image is virtual and diminished

Answer 2

Answer:

Given : focal length=f=-10cm ,object distance =u=-30cm ,height of the image =h2=-5cm ,

image distance =v=? height of the object =h=?

According to the mirror formuka

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \frac{1}{v} = \frac{1}{ - 10} - \frac{1}{ - 30} \\ \frac{1}{v} = \frac{ - 3 + 1}{30} \\ \frac{1}{v} = \frac{ - 2}{30} \\ \frac{1}{v} = \frac{ - 2}{30} \\ \frac{1}{v} = \frac{1}{ - 15} cm \: \\ v = - 15$

Magnification

$m = \frac{h2}{h1} = - \frac{v}{u} \\ h1 = \frac{uh2}{v} \\ h1 = - \frac{ - 30 \times - 5}{ - 15} \\ h1 = - 2 \times - 5 \\ h1 = 10cm \:$

Rajshree has to place the screen 15 cm to the left of the mirror

The height of the object is 1p cm. this the image will be real and diminished.