Question

An object is placed 16.0 cm in front of a concave mirror of focal length 14.0 cm. The mirror forms a real
and inverted image of height 17.5 cm.
What is the height of the object, ho?

please only answer!!​

Answer 1

Answer:-

Height of the object is 2.5 cm .

Explanation:-

We have :-

→ Distance of the object = 16 cm

→ Focal length of the mirror = 14 cm

→ Height of the image = -17.5 cm

[As image is inverted]

________________________________

Since the mirror is concave, we have :-

• u = -16 cm

• f = -14 cm

According to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/(-14) - 1/(-16)

⇒ 1/v = (-1)/14 - (-1)/16

⇒ 1/v = (-8 + 7)/112

⇒ 1/v = -1/112

⇒ v = -112 cm

________________________________

Now, we know that :-

m = -(v/u) = hₑ/hₒ

⇒ -(v/u) = hₑ/hₒ

⇒ -[(-112)/(-16)] = 17.5/hₒ

⇒ -7 = -17.5/hₒ

⇒ hₒ = -17.5/-7

⇒ hₒ = 2.5 cm

Answer 2

${\large{\pmb{\sf{\underline{QuEstion...}}}}}$

★ An object is placed 16.0 cm in front of a concave mirror of focal length 14.0 cm. The mirror forms a real and inverted image of height 17.5 cm. What is the height of the object, h_o?

${\large{\pmb{\sf{\underline{GivEn \; that...}}}}}$

★ An object is placed 16.0 cm in front of a concave mirror of focal length 14.0 cm. The mirror forms a real and inverted image of height 17.5 cm

${\large{\pmb{\sf{\underline{To \: fiNd...}}}}}$

★ The height of the object, hₒ

${\large{\pmb{\sf{\underline{SolutioN...}}}}}$

★ The height of the object, hₒ = 2.5 cm

${\large{\pmb{\sf{\underline{UsiNg \; formulas...}}}}}$

• Formula to find out the magnification but have to work according to the question =

• m = -(v/u) = hₑ/hₒ

• Mirror formula =

• 1/v + 1/u = 1/f

${\large{\pmb{\sf{\underline{Full \; SolutioN...}}}}}$

~ Firstly we have to use formula to find out the magnification but we have to work according to the question

➢️ 1/v + 1/u = 1/f

➢️ 1/v = 1/f - 1/u

➢️ 1/v = 1/(-14)-1/(-16)

➢️ 1/v = 1/-14 - (-1)/16

➢ ️1/v = (-8+7)/112

➢️ 1/v = -1/112

➢️ v = -112 cm

~ Now let's find the height by using the formula to find out the magnification

➢️ m = -(v/u) = hₑ/hₒ

➢️ m = -(-112/-16) = 17.5 / hₒ

➢️ -7 = 17.5/hₒ

➢️ -7/-17.5 = hₒ

➢️ 2.5 = hₒ

➢ hₒ = 2.5 cm

Henceforth, the height of the object, hₒ is 2.5 centimetres.

${\large{\pmb{\sf{\underline{Additional \; KnowlEdge...}}}}}$

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.

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