Question

A 2.0cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 10 cm. The distance of the object from the mirror is 15 cm. Find the nature, position and size of the image formed. Represent the situation with the help of a ray diagram.

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Answer 1

Given↝

f = -10 cm

u = -15 cm

h = 2.0 cm

To find↝

• Find the nature, position and size of the image formed. Represent the situation with the help of a ray diagram.

Using the mirror formula↝

$\: \: \: \: \: \: \: \implies\tt\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\: \: \: \: \: \: \: \: \implies \tt\frac{1}{v} + \frac{1}{ - 15} = \frac{1}{ - 10}$

$\: \: \: \: \: \: \: \implies\tt \frac{1}{v} = \frac{1}{15} - \frac{1}{10}$

$\implies \tt \purple{v = - 30.0 \: cm}$

The image is formed at a distance of 30cm in front of the mirror. Negative sign shows that the image formed is real and inverted.

Magnification↝

$\: \: \: \: \: \: \: \: \: \: \: \sf \: m = \frac{h {}^{1} }{h} = \frac{ - v}{u}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf {h}^{1} = h \times \frac{v}{u} = - 2 \times \frac{ - 30}{ - 15}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{h}^{1} = - 4 \: cm}$

Hence, the size of image is 4cm. Thus, image formed is real, inverted and enlarged.

Some more information↝

• The surface may be either convex or concave.

• Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

Answer 2

★ Question :-

A 2.0cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 10 cm. The distance of the object from the mirror is 15 cm. Find the nature, position and size of the image formed. Represent the situation with the help of a ray diagram.

★ Given that :-

• f = -10 cm

• u = -15 cm

• h = 2.0 cm

★ To find :-

• Find the nature, position and size of the image formed.
• Represent the situation with the help of a ray diagram.

★ Mirror formula :-

$\: \: \: \: \: \red{\star\implies}\bf\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\: \: \: \: \: \: \red{ \star\implies} \bf\frac{1}{v} + \frac{1}{ - 15} = \frac{1}{ - 10}$

$\: \: \: \: \: \red{\star \implies}\bf \frac{1}{v} = \frac{1}{15} - \frac{1}{10}$

$\bigstar \: \boxed{\boxed{\bf\blue{v = - 30.0 \: cm}}}$

The image is formed at a distance of 30cm in front of the mirror. Negative sign shows that the image formed is real and inverted.

★ Magnification :-

$\pink \bigstar \: \bf \: m = \frac{h {}^{1} }{h} = \frac{ - v}{u}$

$\pink\bigstar \: \bf {h}^{1} = h \times \frac{v}{u} = - 2 \times \frac{ - 30}{ - 15}$

$\: \: \: \: \: \: \: \bigstar \: \boxed{ \boxed{ \boxed{\bf \red{{h}^{1} = - 4 \: cm}}}}$

Therefore, the size of image is 4cm and the image formed is real, inverted and enlarged.