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Solution:

Given :

An object 3.5 cm in length which means

[tex] \sf{}Height \: of \: the \: object = 3.5 \: cm[/tex]

Now, Since the mirror used for reflection is converging mirror . According to sign convention for a converging mirror (concave mirror) :

Height of object = Positive

Focal length = Negative

Image Distance= either positive or negative

Object distance = Always negative

Now , Object distance = - 27cm

Focal length = -18cm

To Find :

(i) Image Distance

(ii) Nature of the image

(iii) size of the image ( height )

Solution:

[tex]\:\:\:\:\:\:\star:\boxed{\sf{}\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}\star[/tex]

[tex]\sf{} \implies \frac{1}{ - 18 } = \frac{1}{v} + \frac{1}{ - 27} \\ \\ \sf{} \implies \frac{1}{ - 18} - \frac{1}{ - 27} = \frac{1}{v} \\ \\ \sf{} \implies \frac{1}{ - 18} + \frac{1}{27} = \frac{1}{v} \\ \\ \: \sf{} \implies \frac{ - 3 + 2}{54} = \frac{1}{v} \\ \\ \sf{} \implies \frac{1}{v} = \frac{ - 1}{54} \\ \\ \sf{} \implies v = - 54cm[/tex]

Therefore, image distance is equal to -54 cm

Here, negative sign of the image represents that image is Inverted and therefore Real too .

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \boxed{ \sf{} m = \frac{v}{u} } \star \\ \\ \sf{}m = \frac{ - ( - 54)}{ - 27} \\ \\ \sf{}m = \frac{54}{27} \\ \\ \sf{}m = - 2[/tex]

By using the rate of magnification we can find height of the image .

[tex]\sf{}m = \frac{height \: of \: the \: image}{height \: of \: the \: object} \\ \\ \sf{} - 2 = \frac{h_{image}}{3.5} \\ \\ \sf{}{h_{image}} = - 2(3.5) \\ \\ \sf{}{h_{image}} = - 7cm[/tex]

Therefore , height of the image is -7cm

From above all calculations we can conclude the nature if the image .

As from the negative sign of m we can conclude that image formed is real and inverted and since m<1 , The image formed is enlarged in size as compared to the size of object .

Therefore ,

Nature of Image ;

(i) Enlarged

(ii) Real

(iii) Inverted

(iv) Beyond C

Explanation:

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