Question

I have asked this q. thrice but
: ( i got cringy answers. Please is there anybody who can give answer right and seriously. It is urgent.​

Answer 1

Explanation:

.Given that,

Object distance = -60 cm

Focal length = 30 cm

(a). We need to calculate the image distance

Using formula of focal length

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

v

1

+

u

1

=

f

1

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

v

1

=

f

1

−

u

1

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{-30}+\dfrac{1}{60}

v

1

=

−30

1

+

60

1

\dfrac{1}{v}=-\dfrac{1}{60}

v

1

=−

60

1

v =-60\ cmv=−60 cm

The image is formed at 60 cm distance from the lens.

The size of the image is equal to the size of object.

(b). We need to calculate the magnification

Using formula of magnification

m=-\dfrac{v}{u}m=−

u

v

Put the value into the formula

m=-\dfrac{-60}{-60}m=−

−60

−60

m=-1m=−1

The nature of the image is real and inverted

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Hence, This is the required solution.

Answer 2

Given— distance of object from mirror- 30cm
Focal length = -60cm
We know that
Distance of object from mirror = u
Focal length = f
1/v + 1/u = 1/f
So, 1/v = 1/f - 1/u
1/v = 1/30 - (-1/60)
Cutting the minus sign we get
1/v = 1+2/60
1/v = 3/60
1/v = 1/20
v = 20cm
Now we know that magnification of mirror is
M = -v/u
M = -20/-60
Cutting the minus sign we get
M = 20/60
M = 1/3

From this the image formed will be diminished, erect and virtual.
That was the answer
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