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 be2 cm3 cm4 cm5 cm

Answer 1

Answer:

• Solution is -5 cm

Explanation:

Provided that:

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

• According to sign convention:
• According to sign convention:Height of the object = + 5 cm
• According to sign convention:Height of the object = + 5 cmFocal length = -f
• According to sign convention:Height of the object = + 5 cmFocal length = -fDistance = -2f

To determine:

• Height of the image

Solution:

• Height of the image = -5 cm

Using concepts:

• Mirror formula
• Magnification formula (mirror)

Using formulas:

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!

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

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

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

• 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.”