A convex mirror used a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 20m from t

2 thoughts on “A convex mirror used a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 20m from t”

1. Given :

   In convex mirror,

   Radius of curvature = 3m

   Object distance = 20 m

   To find :

   The position of the image and nature of the image.

   Solution :

   1st we have to find focal length we know that,

   » For a spherical mirror having small aperture, the principle focus lies exactly mid way between the pole and centre of curvature. So, the focal length of a spherical mirror is equal to the half of its radius of curvature.

   if f is the focal length of a mirror and R is its radius of curvature, then f = R/2

   by substituting the given values in the formula,

   [tex]\dashrightarrow \sf f = \dfrac{R}{2}[/tex]

   [tex]\dashrightarrow \sf f = \dfrac{3}{2}[/tex]

   [tex]\dashrightarrow \sf f = 1.5 \: m[/tex]

   Now, using mirror formula that is,

   » A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

   [tex]\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }[/tex]

   where,

   • v denotes Image distance
   • u denotes object distance
   • f denotes focal length

   By substituting all the given values in the formula,

   [tex]\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ – 20} = \dfrac{1}{1.5}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{v} – \dfrac{1}{ 20} = \dfrac{1}{1.5}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{ 20}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{v} = \dfrac{40 + 3}{ 60}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{v} = \dfrac{43}{ 60}[/tex]

   [tex]\dashrightarrow\sf v = \dfrac{60}{43}[/tex]

   [tex]\dashrightarrow\sf v = 1.39 \: m[/tex]

   Thus, the position of the image is 1.39 m.

   Hence, the nature of image is the image is erect and virtual.

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2. Hola ⚘⚘

   [tex]Given :-[/tex]

   • Radius of curvature = 3m
   • Object distance = 20 m

   [tex]To find :-[/tex]

   • Position of the image
   • Nature of the image.

   [tex]Solution :-[/tex]

   Finding the focal length

   [tex]\implies\:\: f = \dfrac{R}{2}[/tex]

   [tex]\implies\:\: f = \dfrac{3}{2}[/tex]

   [tex]\implies\:\:f = 1.5 \: m[/tex]

   Now, Using mirror formula

   [tex]\implies\:\: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}[/tex]

   v = Image distance

   u = object distance

   f = focal length

   [tex]\implies\:\: \dfrac{1}{v} + \dfrac{1}{ – 20} = \dfrac{1}{1.5}[/tex]

   [tex]\implies\:\: \dfrac{1}{v} – \dfrac{1}{ 20} = \dfrac{1}{1.5}[/tex]

   [tex]\implies\:\: \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{ 20}[/tex]

   [tex]\implies\:\: \dfrac{1}{v} = \dfrac{40 + 3}{ 60}[/tex]

   [tex]\implies\:\: \dfrac{1}{v} = \dfrac{43}{ 60}[/tex]

   [tex]\implies\:\: v = \dfrac{60}{43}[/tex]

   [tex]\implies\:\: v = 1.39 \: m[/tex]

   • Position of the image is 1.39 m

   • Nature of image erect and virtual

   Reply