Question

Please answer ASAP
An object of size 7.5 cm is kept at a distance of 35 cm from a conve lens of focal length 15 cm. Choose a suitable scale to locate the position, nature and size of
the image formed by the lens.

Answer 1

$\bf{\underline{\underline{Given}}}$

$\mathsf{\bigstar\: height\:of\:object \: (h_{o}) = 7.5\:cm}$

$\mathsf{\bigstar\: object \: distance \: (u) = -35\:cm}$

$\mathsf{\bigstar\: focal \: length \: (f) = 15\:cm}$

$\bf{\underline{\underline{To\:find}}}$

$\mathsf{\bigstar\: Image\:distance\: (v) }$

$\mathsf{\bigstar\: Nature \: of \: the \: Image}$

$\mathsf{\bigstar\: Size\: of \: the \: Image\: (h_{i})}$

$\bf{\underline{\underline{Solution}}}$

$\fbox{\red{\mathsf{Len's\: formula : \: \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}}$

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

$\mathsf{\implies \: \frac{1}{v} = \frac{1}{15} + \frac{1}{-35}}$

$\mathsf{\implies \: \frac{1}{v} = \frac{1}{15} - \frac{1}{35}}$

$\mathsf{\implies \: \frac{1}{v} = \frac{7 - 3}{105}}$

$\mathsf{\implies \: \frac{1}{v} = \frac{4}{105}}$

$\mathsf{\implies \: v = \frac{105}{4}}$

$\mathsf{\implies \: v = 26.25 \: cm}$

$\fbox{\mathsf{\green{Magnification = \frac{h_{i}}{h_{o}} = \frac{v}{u}}}}$

$\mathsf{\implies \: \frac{h_{i}}{7.5} = \frac{26.25}{-35}}$

$\mathsf{\implies \: h_{i} = \frac{26.25 × 7.5}{-35}}$

$\mathsf{\implies \: h_{i} = -5.625 \: cm}$

∴ The image formed is real, inverted, diminshed

The image is formed between F and 2F (On the opposite side)

Answer 2

Answer:

Mirror formula :

v

1

+

u

1

=

f

1

...........(1)

So, u=−27cm

Concave mirror have negative focal length.

f=−18cm

Putting the values in (1),

v

1

+

−27

1

=

−18

1

We get , v=−54cm

Also,

Magnification =

h

1

h

2

=

u

−v

7.0

h

2

=−

−27

−54

h

2

=−14.0cm

Image will be real and inverted and will be of size 14 cm.