Question

(b) Focal length of a concave mirror is 20 cm. What is the nature of image if an
object is placed at 60 cm?​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\implies\textsf{ Focal length of the concave mirror is 20cm. }\\\implies\textsf{ Object is placed at 60 cm .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The nature of the image .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

We know that the radius of curvature is double of focal length ( for small apertures ). So here the focal length is 20 cm. So the radius of curvature will be 20cm * 2 = 40 cm . And the object is placed at 60cm. So the object is placed beyond C. In this case , the image is , formed between centre of curvature and the focus . The size of the image is smaller than the size of the object and the image is real and inverted . [ Also refer to attachment : ] Now , according to the given data , let's calculate the image distance , whether the given nature of image is satisfied .

$\underline{\green{ \sf Nature \ of \ the \ image :- }}$

$\boxed{\begin{array}{r|r} \underline{\red{\sf Position }} & \sf Between \ C \ and \ F \\\\ \underline{\red{\sf Nature }} &\sf Real \ and \ inverted \\\\ \underline{\red{\sf Size }} & \textsf{ \sf Size of {image} < Size of object}\\\end{array} }$

$\underline{\purple{\boldsymbol{ Using \ Mirror \ Formula ,\ we \ have :- }}}$

$\sf:\implies \pink{\dfrac{1}{Focal \ lenght (f) } = \dfrac{ 1}{Object \ Distance ( u) }+\dfrac{1}{Image \ Distance (v) }} \\\\\sf:\implies \dfrac{1}{-20cm} = \dfrac{1}{-60cm} + \dfrac{1}{v}\\\\\sf:\implies \dfrac{1}{v} = \dfrac{1}{60cm}-\dfrac{1}{20cm} \\\\\sf:\implies \dfrac{1}{v} = \dfrac{ 1-3}{60cm}\\\\\sf:\implies\dfrac{1}{v}=\dfrac{-2}{60cm} \\\\\sf:\implies \dfrac{1}{v}=\dfrac{-1}{30cm} \\\\\sf:\implies\boxed{\pink{\mathfrak{ Image \ distance = -30 cm }}}$

$\underline{\blue{\sf Hence \ the \ Image \ distance\ is \ \textsf{\textbf{-30cm }}. }}$

Hence here we can see that the image is formed at 30cm from the pole , which is between the focal length ( 20cm) and the centre of curvature ( 40cm ) . Hence the stated characteristics are correct .

$\rule{200}2$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Some \ more \ Information :- }}}$

• Sign Convention :-

$\blue{\longmapsto}$The distances measured along the direction of incident ray are taken as positive and that opposite to the direction of incident rays are taken as negative.

$\blue{\longmapsto}$ All the distances parallel to the principal axis are measured from the optical centre.

$\blue{\longmapsto}$All the distances measured perpendicular to and above the principal axis are taken positive while those measured below it are taken as negative.

$\blue{\longmapsto}$Focal length is taken positive for convex lens while negative for concave lens object distance is always taken as negative for all objects.

$\rule{200}2$