Question

focal length of the concave mirror is 8 cm the object is placed at 16 cm in front of the mirror write the position of the image and its nature with
diagram​

Answer 1

$\longrightarrow$ f = -8 cm (acc to sign convention)

$\longrightarrow$ u = -16 cm

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

$\longrightarrow\frac{1 }{ - 8} = \frac{1}{ - 16} + \frac{1}{v}$

$\longrightarrow\frac{1}{ - 8} + \frac{1}{ 16} = \frac{1}{v}$

$\longrightarrow\frac{1}{v} = \frac{ - 8 + 16}{ - 8 \times 16}$

$\longrightarrow\frac{1}{v} = \frac{8}{ - 128}$

$\longrightarrow\frac{1}{v} = \frac{1}{ - 16}$

$\longrightarrow \: v = - 16 \: cm$

$\longrightarrow$To find the nature of the image we must find the magnification

$\longrightarrow \: m = \frac{ - v}{u}$

$\longrightarrow \: m = \frac{ - ( - 16)}{ - 16}$

$\longrightarrow \: m = - 1$

$\longrightarrow$ As the magnification is less than 1 the nature of the image is real and inverted and the size of image is diminished