Question

A 7.0 cm tall light bulb is placed a distance of 37.5 cm from a convex mirror having a focal length of -12.5 cm. Determine the image distance and the image size.

Answer 1

Answer ⤵️

We have,

Object placed, distance (u) = -27 cm

A concave mirror of focal length (f) = -18 cm

We know, 1/f = 1/u + 1/v

or, 1/(-18) = 1/(-27) + 1/v

or, 1/v = 1/18 + 1/27

or, v = - 54 cm

Hence image formed at distance -54 cm

we know,

Magnification (m ) = - image distance (v)/ object distance(u) = Image height (h2)/ Object height (h1)

where, object height (h1) = 7 cm

Therefore,

- (54)/(-27) = Image height (h2)/7

or, Image height (h2) = - (54) x 7/(-27) =-14 cm

So, Image is formed on the same side as that of the object, since, negative sign indicated. The size of the image is 14 cm and inverted.

Answer 2

Answer:

The image distance and the image size are -18.57 cm and 3.4664 cm respectively.

Explanation:

From the mirror formula we have,

$\mathrm{\frac{1}{f}= \frac{1}{v}+ \frac{1}{u} }$         (1)

Where,

f=focal length of the mirror

v=image distance from the mirror

u=object distance from the mirror

From the question we have,

The height of the object(h₁)=7 cm

The object distance from the mirror(v)=-37.5 cm

The focal length of the mirror(f)=-12.5cm

By inserting the value of "v" and "f" in equation (1) we get;

$\mathrm{\frac{1}{-12.5}= \frac{1}{v}+ \frac{1}{-37.5} }$

$\mathrm{\frac{1}{v}=\frac{-1}{12.5} +\frac{1}{37.5} }$

$\mathrm{\frac{1}{v}=\frac{-3+1}{37.5} }$

$\mathrm{\frac{1}{v}=\frac{-2}{37.5} }$

$\mathrm{v=\frac{-37.5}{2} }$

$\mathrm{v=-18.57\ cm}$          (2)

And the magnification of the mirror is calculated as,

$\mathrm{m=\frac{-v}{u} }$         (3)

$\mathrm{m=\frac{h_2}{h_1} }$         (4)

h₂=image height

h₁=object height

From equations (3) and (4) we get;

$\mathrm{\frac{-v}{u} =\frac{h_2}{h_1} }$         (5)

By inserting the value of "v", "u" and "h₁" in equation (5) we get;

$\mathrm{\frac{-(-18.57)}{-37.5} =\frac{h_2}{7} }$

$\mathrm{\frac{h_2}{7}=0.4952 }$

$\mathrm{h_2=3.4664\ cm}$       (6)

So, the image distance and the image size are -18.57 cm and 3.4664 cm respectively.

#SPJ3