Question

the radius of curvature of a convex mirror used in an automobile is 200 CM a bus is Coming from behind it at a distance of 3.5 metre calculate the position and size of image relative to but also comment on the nature of image calculate all calculation in metre​

Answer 1

Answer:

1. Position of image is 0.77 m.
2. The size of image is 0.22 times that of object.
3. Nature of image is virtual, erect and diminished.

Explanation:

Given that:

• The radius of curvature of a convex mirror used in an automobile is 200 cm.
• A bus is coming from behind it at a distance of 3.5 metre.

To Find:

1. Position of image.

2. Relative size of image.

• i.e., Magnification.

3. Nature of image.

Formula used:

Mirror formula.

1. 1/v + 1/u = 1/f

2. m = - v/u

Where,

• Image distance is denoted as v.
• Object distance is denoted as u.
• Focal length is denoted as f.
• Magnification is denoted as m.

We have:

• Focal length, f = R/2 = 200/2
• f = 100 cm = 1 m [Focal Length is taken positive in convex mirror]
• Object distance, u = - 3.5 m [In front of the mirror]

Finding the position of image:

⇒ 1/v + 1/u = 1/f

• Substituting the values.

⇒ 1/v - 1/3.5 = 1/1

⇒ 1/v = 1/1 + 1/3.5

⇒ 1/v = 3.5/3.5 + 1/3.5

⇒ 1/v = 4.5/3.5

⇒ v = 3.5/4.5

⇒ v = 0.77 (approx.)

∴ Position of image = 0.77 m

Finding the relative size of the image:

⇒ m = - v/u

• Substituting the values.

⇒ m = - 0.77/(- 3.5)

⇒ m = 0.77/3.5

• Minus sign cancelled.

⇒ m = 0.22

∴ Magnification = 0.22

Size of image = 0.22 times of object

Finding the nature of image:

• The value of the position of image is positive i.e., + 0.77 m. We can say that image is formed behind the mirror.
• So, virtual and erect image is formed
• And also we get the magnification as 0.22.
• It means diminished image is formed.

Answer 2

${\large{\frak{\pmb{\underline{Given \; that}}}}}$

★ The radius of curvature of a convex mirror used in an automobile is 200 cm.

★ A bus is coming from behind it at a distance of 3.5 metre.

${\large{\frak{\pmb{\underline{To \; determine}}}}}$

★ The position of image relative to bus.

★ The size of image relative to bus.

★ The nature of image.

${\large{\frak{\pmb{\underline{Note}}}}}$

★ Calculate all calculation in metres.

${\large{\frak{\pmb{\underline{Solution}}}}}$

★ The position of image relative to bus = 0.77 metres

★ The size of image relative to bus = 0.22

★ The nature of image = Erect, virtual and diminished.

${\large{\frak{\pmb{\underline{Using \; dimensions}}}}}$

★ Mirror formula.

★ Magnification formula.

${\large{\frak{\pmb{\underline{Using \; formula}}}}}$

★ 1/v + 1/u = 1/f

★ m = -v/u

${\large{\frak{\pmb{\underline{Where,}}}}}$

★ v denotes distance of the image.

★ u denotes distance of the object.

★ f denotes focal length.

★ m denotes magnification

${\large{\frak{\pmb{\underline{Here,}}}}}$

★ Distance of the image, v is ?

★ Distance of the object, u is -3.5 metres (In front it the mirror)

★ Focal length, f is 1 m (how?)

➣ Focal length = Radius/2

➣ Focal length = 200/2

➣ Focal length = 100 centimetres

➣ 1 m = 100 cm

➣ 1 metres

★ Magnification, m is ?

${\large{\frak{\pmb{\underline{Full \; Solution}}}}}$

☀ Firstly let us find the position of image relative to bus.

➟ 1/v + 1/u = 1/f

➟ 1/v + 1/(-3.5) = 1/1

➟ 1/v - 1/3.5 = 1/1

➟ 1/v = 1/1 + 1/3.5

(Let's find LCM and let's do)

➟ 1/v = (3.5+1)/3.5

➟ 1/v = 4.5/3.5

➟ v = 0.77 metres (approx)

• Henceforth, 0.77 metres is the position of image relative to bus.

☀ Now let us find the size of image relative to bus / magnification.

➟ m = -v/u

➟ m = -0.77/(-3.5)

➟ m = 0.77/3.5

➟ m = 0.22

• Henceforth, 0.22 is the size of image relative to bus / magnification.

☀ Now let's find the nature of the image.

➟ As 0.77 metres is the position of image relative to bus. Henceforth, it is placed behind the mirror. Means it's an erect and virtual image.

➟ It's magnification is 0.22 henceforth, it is diminished too.

${\large{\frak{\pmb{\underline{Additional \; knowledge}}}}}$

Ray diagram convex mirror.

$\setlength{\unitlength}{0.7 cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){12}}\qbezier(10.49,0)(10.5,1.8)(8.5,3.8)\qbezier(10.49,0)(10.5,-1.8)(8.5,-3.8)\put(7,0){\circle*{0.2}}\put(4,0){\circle*{0.2}}\put(2,0){\vector(0,1){1.5}}\linethickness{0.1mm}\put(2,1.5){\line(1,0){8.2}}\qbezier(10.2,1.5)(7,0)(3.8,-1.5)\put(2,1.5){\line(4,-3){6.77}}\thicklines\put(5.17,0){\vector(0,-1){0.85}}\put(6,1.496){\vector(1,0){0}}\put(6.3,1.496){\vector(1,0){0}}\put(4.2,-1.33){\vector(-3,-2){0}}\put(8,-2.96){\vector(3,-2){0}}\put(7.5,-2.64){\vector(-3,2){0}}\put(1.9,-0.5){\sf B}\put(1.9,1.7){\sf A}\put(3.7,-0.5){\sf C}\put(7.2,-0.5){\sf F}\put(5,-1.4){\sf A'}\put(5,0.2){\sf B'}\put(10.7,0.2){\sf P}\put(10.35,1.45){\sf D}\put(9,-4){\sf E}\end{picture}$

Image formation in Concave mirror.

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{cccc}\sf \pink{Position_{\:(object)}} &\sf \purple{Position_{\:(image)}} &\sf \red{Size_{\:(image)}} &\sf \blue{Nature_{\:(image)}}\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf At \:Infinity &\sf At\: F&\sf Highly\:Diminished&\sf Real\:and\:Inverted\\\\\sf Beyond\:C &\sf Between\:F\:and\:C&\sf Diminished&\sf Real\:and\:Inverted\\\\\sf At\:C &\sf At\:C&\sf Same\:Size&\sf Real\:and\:Inverted\\\\\sf Between\:C\:and\:F&\sf Beyond\:C&\sf Enlarged&\sf Real\:and\;Inverted\\\\\sf At\:F&\sf At\:Infinity&\sf Highly\: Enlarged&\sf Real\:and\:Inverted\\\\\sf Between\:F\:and\:P&\sf Behind\:the\:mirror&\sf Enlarged&\sf Erect\:and\:Virtual\end{array}}\end{gathered}\end{gathered}\end{gathered}$