Question

An object of height 5 cm is placed on the principal axis of a concave mirror of focal length f at a distance of 2f .The height of the image will be
a) 2cm
b) 3cm
c) 4cm
d) 5cm


it's important answer quickly...​

Answer 1

Answer :-

The image formed, is an inverted image of height 5 cm [Option.d]

Explanation :-

We have :-

→ Height of object (hₒ) = 5 cm

→ Focal length of mirror = f

→ Distance of object = 2f

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According to sign convention, we have :-

• u = -2f

• f = -f

Now, we shall put the above values in the mirror formula to obtain the position of the image.

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/(-f) - 1/(-2f)

⇒ 1/v = -1/f + 1/2f

⇒ 1/v = (-2 + 1)/2f

⇒ 1/v = -1/2f

⇒ v = -2f

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Now, we can calculate the height of the image as follows :-

m = -(v/u) = hᵢ/hₒ

⇒ -(-2f/-2f) = hᵢ/5

⇒ -1 = hᵢ/5

⇒ hᵢ = 5(-1)

⇒ hᵢ = -5 cm

[Here, -ve sign represents inverted image] .

Answer 2

Answer:

Provided that:

• Concave mirror is given.
• Height of object = 5 cm.
• Focal length = f.
• Distance = 2f.

According to sign convention:

• Height of the object = + 5 cm
• Focal length = -f
• Distance = -2f

To determine:

• Height of the image

Solution:

• Height of the image = -5 cm

Using concepts:

• Mirror formula
• Magnification formula (mirror)

Using formulas:

✡️ Mirror formula is mentioned:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{f} \: = \dfrac{1}{u} + \dfrac{1}{v}}}}}}}$

✡️ Magnification formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{-v}{u} \: = \dfrac{h'}{h}}}}}}}$

Where, v denotes image distance, u denotes object distance, f denotes focal length, h′ denotes height of the image and h denotes height of the object.

Required solution:

✴️ Firstly by using mirror formula let us find out the value of image distance, v!

$:\implies \tt \dfrac{1}{f} \: = \dfrac{1}{u} + \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{1}{-f} = \dfrac{1}{-2f} + \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{1}{-f} = - \dfrac{1}{2f} + \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{1}{-f} + \dfrac{1}{2f} = \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{-f \times 1 + 1 \times 1}{2f} \: = \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{-f + 1}{2f} \: = \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{-f}{2f} \: = \dfrac{1}{v} \\ \\ :\implies \tt \dfrac{-1}{2f} \: = \dfrac{1}{v} \\ \\ :\implies \tt -v \: = 2f \\ \\ :\implies \tt v \: = -2f \: cm \\ \\ :\implies \tt Distance_{image} \: = -2f \: cm$

✴️ Now by using suitable magnification formula let us find out the height of the image!

$:\implies \tt \dfrac{-v}{u} \: = \dfrac{h'}{h} \\ \\ :\implies \tt \dfrac{-(-2f)}{-2f} = \dfrac{h'}{5} \\ \\ :\implies \tt \dfrac{2f}{-2f} = \dfrac{h'}{5} \\ \\ :\implies \tt \dfrac{1}{-1} = \dfrac{h'}{5} \\ \\ :\implies \tt 1 \times 5 = -1 \times h' \\ \\ :\implies \tt 5 = -h' \\ \\ :\implies \tt h' \: = -5 \: cm \\ \\ :\implies \tt Height_{image} \: = -5 \: cm$

• The nature of the image is real and inverted.

Additional information:

• If the magnification produced by a spherical mirror is in negative then the mirror is always “Concave Mirror.”

• If the magnification produced by a spherical mirror is in positive then the mirror is always “Convex Mirror.”

• If magnification is negative in a concave mirror then it's nature is “Real and Inverted” always.

• If magnification is positive in a convex mirror then it's nature is “Virtual and Erect” always.

• If in the ± magnification, magnitude > 1 then the image formed is “Enlarged”.

• If in the ± magnification, magnitude < 1 then the image formed is “Diminished”.

• If in the ± magnification, magnitude = 1 then the image formed is “Same sized”.

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.”