Question

a concave mirror has a focal length of 4cm and an object 2cm tall is placed 9cm away from if find the nature position and size of the image formed​

Answer 1

The object is Real, Inverted diminished and F & 2F with a height of -1.6

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$\huge\gray{\textsf{\textbf Given}}$

• Mirror = Concave
• Focal Length = -4 cm
• Height of image = 2 cm
• Object Distance = -9 cm = u

$\huge\gray{\textsf{\textbf To Find}}$

• Nature, Position and Size of the image

$\huge\gray{\textsf{\textbf Solution}}$

✭ Image Distance

$\sf \star \: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\sf \dashrightarrow \dfrac{1}{v} + \dfrac{1}{-9} = \dfrac{1}{-4}$

$\sf \dashrightarrow \dfrac{1}{v} - \dfrac{1}{9} = -\dfrac{1}{4}$

$\sf \dashrightarrow \dfrac{1}{v} = \dfrac{1}{9} - \dfrac{1}{4}$

$\sf \dashrightarrow \dfrac{1}{v} = \dfrac{4}{36} - \dfrac{9}{36}$

$\sf \dashrightarrow \dfrac{1}{v} = \dfrac{4-9}{36}$

$\sf \dashrightarrow \dfrac{1}{v} = \dfrac{-5}{36}$

$\sf \dashrightarrow v = \dfrac{36}{-5}$

$\sf \dashrightarrow\purple{v = -7.2}$

✭ Magnification

$\sf \star \: Magnification = \dfrac{-v}{u}$

$\sf \dashrightarrow M = \dfrac{-(-7.2)}{-9}$

$\sf \dashrightarrow M = \dfrac{7.2}{9}$

$\sf \dashrightarrow \red{M = -0.8}$

✭ Height of the image

$\sf \star \: M = \dfrac{h_i}{h_o}$

$\sf \dashrightarrow -0.8 = \dfrac{h_i}{2}$

$\sf \dashrightarrow -0.8\times 2 = h_i$

$\sf \dashrightarrow \pink{h_i = -1.6}$