Question

Please solve the above question!!

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Answer 1

Answer:

Given

 Size of the object [o] = +7cm

  F = -18cm

 U= -27cm

 V = ?

 1/f = 1/v+1/u

 

1/-18cm = 1/v-1/27

1/v= -1/54 v = -54cm

Since Vis – ve screen should be placed at 54cm in front of the mirror.

Size of the image : m= I/O= -v/u I/7=-[54/27] I= -14cm

Nature of the image : image is real , inverted and magnified. 

Answer 2

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Here,

object distance, u=-27 cm (to left side of mirror)

Image distance, v =? (to be calculated)

As we know it is a concave mirror,

So, focal length, f=-18 cm

Height of object, h1 =7.0 cm

Height of image, h2 =? (to be calculated)

Now for a mirror :

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} + \frac{1}{ - 27} = \frac{1}{ - 18} \\ \frac{1}{v} - \frac{1}{27} = - \frac{1}{18} \\ \frac{1}{v} = - \frac{1}{18} + \frac{1}{27} \\ \frac{1}{v} = \frac{ - 3 + 2}{54} \\ \frac{1}{v} = - \frac{1}{54}$

Image distance, v=-54 cm

The screen should be placed at a distance of 54 cm in front of the concave mirror on its left side.

Also, for a mirror,

$\frac{h2}{h1} = - \frac{v}{u} \\ \frac{h2}{7.0} = \frac{ - 54}{ - 27} \\ \frac{h2}{7.0} = - 2 \\ h2 = - 2 \times 7.0 \\ h2 = - 14.cm$

Thus, the size of image is 14 cm.