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

Question

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 this position of image. What is the nature of image ?

in progress 0
Adeline 2 years 2021-06-22T02:48:10+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-06-22T02:49:13+00:00

    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,

    \dashrightarrow \sf f = \dfrac{R}{2}

    \dashrightarrow \sf f = \dfrac{3}{2}

    \dashrightarrow \sf f = 1.5 \: m

    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.,

    \boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }

    where,

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

    By substituting all the given values in the formula,

    \dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

    \dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{1.5}

    \dashrightarrow\sf \dfrac{1}{v}  -  \dfrac{1}{ 20} = \dfrac{1}{1.5}

    \dashrightarrow\sf \dfrac{1}{v}   = \dfrac{1}{1.5} + \dfrac{1}{ 20}

    \dashrightarrow\sf \dfrac{1}{v}   =  \dfrac{40  + 3}{ 60}

    \dashrightarrow\sf \dfrac{1}{v}   =  \dfrac{43}{ 60}

    \dashrightarrow\sf v   =  \dfrac{60}{43}

    \dashrightarrow\sf v   =  1.39 \: m

    Thus, the position of the image is 1.39 m.

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

    0
    2021-06-22T02:49:14+00:00

    Hola ⚘⚘

    Given :-

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

    To find :-

    • Position of the image
    • Nature of the image.

    Solution :-

    Finding the focal length

    \implies\:\: f = \dfrac{R}{2}

    \implies\:\: f = \dfrac{3}{2}

    \implies\:\:f = 1.5 \: m

    Now, Using mirror formula

    \implies\:\: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

    v = Image distance

    u = object distance

    f = focal length

    \implies\:\: \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{1.5}

    \implies\:\: \dfrac{1}{v}  -  \dfrac{1}{ 20} = \dfrac{1}{1.5}

    \implies\:\: \dfrac{1}{v}   = \dfrac{1}{1.5} + \dfrac{1}{ 20}

    \implies\:\: \dfrac{1}{v}   =  \dfrac{40  + 3}{ 60}

    \implies\:\: \dfrac{1}{v}   =  \dfrac{43}{ 60}

    \implies\:\: v   =  \dfrac{60}{43}

    \implies\:\: v   =  1.39 \: m

    • Position of the image is 1.39 m

    • Nature of image erect and virtual

Leave an answer

Browse

9:3-3+1x3-4:2 = ? ( )