An object 2cm high is placed at a distance of 16 cm from a concave mirror which produces 3cm high inverted image find the focal le

2 thoughts on “An object 2cm high is placed at a distance of 16 cm from a concave mirror which produces 3cm high inverted image find the focal le”

1. Merhabalar ⚘⚘

   [tex]Given :-[/tex]

   • Height of the object = 2 cm
   • Object distance = -16 cm
   • Height of the image = 3 cm

   [tex]To find :-[/tex]

   • Focal length of the object

   [tex]Solution :-[/tex]

   Finding the image distance,

   [tex]:\implies\:\: \dfrac{h’}{h} = – \dfrac{v}{u} [/tex]

   Where,

   • h’ = height of the image
   • h = height of the object
   • v = image distance
   • u = object distance

   [tex]:\implies\:\: \dfrac{3}{2} = – \dfrac{v}{ – 16} [/tex]

   [tex]:\implies\:\: v = \dfrac{16 \times 3}{2} [/tex]

   [tex]:\implies\:\: v = 8 \times 3[/tex]

   [tex]:\implies\:\: v = 24 \: cm[/tex]

   Now, using mirror formula that is,

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

   Where,

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

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

   [tex]:\implies\:\: \dfrac{1}{ – 24} + \dfrac{1}{ – 16} = \dfrac{1}{f}[/tex]

   [tex]:\implies\:\: \dfrac{1}{ 24} – \dfrac{1}{16} = \dfrac{1}{f}[/tex]

   [tex]:\implies\:\: \dfrac{1}{f} = \dfrac{2 – 3}{48}[/tex]

   [tex]:\implies\:\: \dfrac{1}{f} = \dfrac{ – 1}{48}[/tex]

   [tex]:\implies\:\: f = – 48 \: cm[/tex]

   • Focal length of the object is -48 cm.

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2. Given :

   In concave mirror,

   Height of the object = 2 cm

   Object distance = – 16 cm

   Height of the image = 3 cm

   To find :

   The focal length of the object

   Solution :

   1st we have to find image distance. We know that,

   » The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

   [tex]\dashrightarrow\bf \dfrac{h’}{h} = – \dfrac{v}{u} [/tex]

   where,

   • h’ denotes height of the image
   • h denotes height of the object
   • v denotes image distance
   • u denotes object distance

   By substituting all the given values in the formula,

   [tex]\dashrightarrow\sf \dfrac{3}{2} = – \dfrac{v}{ – 16} [/tex]

   [tex]\dashrightarrow\sf v = \dfrac{16 \times 3}{2} [/tex]

   [tex]\dashrightarrow\sf v = 8 \times 3[/tex]

   [tex]\dashrightarrow\sf v = 24 \: cm[/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}{ – 24} + \dfrac{1}{ – 16} = \dfrac{1}{f}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{ 24} – \dfrac{1}{16} = \dfrac{1}{f}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{f} = \dfrac{2 – 3}{48}[/tex]

   [tex]\dashrightarrow\sf \dfrac{1}{f} = \dfrac{ – 1}{48}[/tex]

   [tex]\dashrightarrow\sf f = – 48 \: cm[/tex]

   Thus, the focal length of the object is -48 cm.

   Reply