Question

A 5cm high object is placed in front of concave mirror with o radius of curvature 20cm determine the image hight if the object distance is 5cm,10cm,20cm,30cm

Answer 1

GIVEN :

• Height of the object, ho = 5 cm
• The radius of the concave mirror, R = 20 cm
• distance of the object from the mirror, Uo= 5 cm/10cm/30 cm

TO FIND

• Height of the image, hi.

SOLUTION :

We can simply solve this problem as under,

we know, that the focal length of the mirror is half of the radius of curvature.

so,

f = -10cm (as the mirror is concave)

Using the mirror formula,

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

putting the values in the above formula, we get;

(I) When u = 5 cm

1/5 + 1/v = -1/10

1/v = -1/10 + 1/5

1/v = 1/10

v = 10

Distance of the image, v = 10cm

Now, applying the formula to find the magnification of the mirror, we have

M = hi/ho = v/u

= hi/5 = 10 /5

hi = 5×2

hi = 10

hence, HEIGHT OF THE IMAGE = 10cm

(ll) When u = 15cm

putting the values in the mirror formula we get,

1/v + 1/u = 1/f

1/v = -1/10 + 1/5

1/v = -1/30

The distance of the image is -30cm

Now, applying the formula the to find magnificationtion of mirror, we have

Now, applying the formula the to find magnificationtion of mirror, we have M = hi/ho = v/u

= hi/5 = -30/15

hi= 5× -2

hi = -10

Hence, The Height Of Image is -10cm.

(III) when u = 20cm

putting the values in the mirror formula we get,

1/v + 1/u = 1/f

1/v = - 1/10 + 1/20

1/v = - 1/20

v = - 20

Distance of the image v = -20cm

Now, applying the formula to find magnification of mirror, we have

Now, applying the formula to find magnification of mirror, we have M = hi/ho = v/u

M = hi /5 = -20/20

hi = 5× -1

Hence, Height Of The Image is -5cm.

(IV) when u = 30 cm

putting the values in the mirror formula we get,

1/v + 1/u = 1/f

1/v = -1/10 + 1/30

1/v = -2/30

1/v = -1/15

v = -15

Distance of the image = -15cm.

Now, applying the formula to find the magnification of the mirror, we have

of the mirror, we have M = hi/ho = v/u

M= hi/5 = -15/30

hi = 5× -1/2

hi = -2.5

Hence, the height of the image is -2.5cm.

Answer 2

Height of the image when object distance is 5 cm is 10 cm

Height of the image when object distance is 15 cm is -10 cm

Height of the image when object distance is 20 cm is -5 cm

Height of the image when object distance is 30 cm is -2.5 cm

Given : Height of the object is 5 cm

Radius of curvature of the concave mirror is 20 cm

To Find : The height of the image if the object distance is

a) 5 cm   b) 10 cm  c) 20 cm  d) 30 cm

Solution :

It is given that a 5cm high object is placed in front of concave mirror with of radius of curvature 20cm.

Now

focal length of mirror is -10 cm (Focal length of the mirror is half of the radios of curvature and focal length of concave lens is always negative)

We know the mirror formula is

$\frac{1}{v}+\frac{1}{u} = \frac{1}{f}$

where v is the distance of the image from the mirror

u is the distance of the object from the mirror

and f is the  focal the focal length of the mirror

a) object distance is 5 cm

Here u = 5 cm

f = -10 cm

v = ?

putting the values in mirror formula , we get

$\frac{1}{v} = \frac{1}{f} -\frac{1}{u}$

$\frac{1}{v} = -\frac{1}{10} +\frac{1}{5}$

= $\frac{1}{10}$

v = 10 cm

Distance of the image from the mirror is 10 cm

Now we know that

Magnificent (M ) = $\frac{hi}{ho}$ = $-\frac{v}{u}$ (here negative sign means that inverse image will formed)

(We will ignore minus sign in calculation)

Where hi is height of the image

and ho is the height of the image

M = $\frac{10}{5}$ = 2

ho = 5 cm

hi = ?

2 = $\frac{hi}{5}$

hi 10

So height of the image is 10 cm

b) object distance is 15 cm

Here u = 15 cm

f = -10 cm

v = ?

putting the values in mirror formula , we get

$\frac{1}{v} = \frac{1}{f} -\frac{1}{u}$

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

= $-\frac{1}{30}$

v = -30 cm

Distance of the image from the mirror is -30 cm

Now we know that

Magnificent (M ) = $\frac{hi}{ho}$ = $-\frac{v}{u}$ (here negative sign means that inverse image will formed)

(We will ignore minus sign in calculation)

Where hi is height of the image

and ho is the height of the image

M = $-\frac{30}{15}$ = -2

ho = 5 cm

hi = ?

-2 = $\frac{hi}{5}$

hi is -10

So height of the image is -10 cm

c) object distance is 20 cm

Here u = 5 cm

f = -10 cm

v = ?

putting the values in mirror formula , we get

$\frac{1}{v} = \frac{1}{f} -\frac{1}{u}$

$\frac{1}{v} = -\frac{1}{10} +\frac{1}{20}$

= $-\frac{1}{20}$

v = -20 cm

Distance of the image from the mirror is -20 cm

Now we know that

Magnificent (M ) = $\frac{hi}{ho}$ = $-\frac{v}{u}$ (here negative sign means that inverse image will formed)

(We will ignore minus sign in calculation)

Where hi is height of the image

and ho is the height of the image

M = $\frac{-20}{20}$ = -1

ho = 5 cm

hi = ?

-1 = $\frac{hi}{5}$

hi is -5

So height of the image is -5 cm

d) object distance is 30 cm

Here u = 5 cm

f = -10 cm

v = ?

putting the values in mirror formula , we get

$\frac{1}{v} = \frac{1}{f} -\frac{1}{u}$

$\frac{1}{v} = -\frac{1}{10} +\frac{1}{30}$

= $\frac{-2}{30}$

v = -15 cm

Distance of the image from the mirror is -15 cm

Now we know that

Magnificent (M ) = $\frac{hi}{ho}$ = $-\frac{v}{u}$ (here negative sign means that inverse image will formed)

(We will ignore minus sign in calculation)

Where hi is height of the image

and ho is the height of the image

M = $\frac{-15}{30}$ = -0.5

ho = 5 cm

hi = ?

-0.5 = $\frac{hi}{5}$

hi is -2.5

So height of the image is -2.5 cm

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