Question

An object is placed in front of a concave lens of focal length 10cm as

at a distance of 5cm. Find the nature and position of image. Draw a ray diagram to explain it .

​

Answer 1

• Image Position is 3.33 cm

Given:-

Focal length ,f = -10 cm

Object distance ,u = -5 cm

To Find:-

Nature & Position of Image .

Solution:-

As we have to calculate the image position .

Using Lens Formula

•• 1/v - 1/u = 1/f

Substitute the value we get

→ 1/v = 1/f + 1/u

→ 1/v = -1/10 + 1/(-5)

→ 1/v = -1/10 - 1/5

→ 1/v = -1 -2/10

→ 1/v = -3/10

→ v = -10/3

→ v = -3.33 cm

Hence, the image position is 3.33 cm .

Now, Magnification ,

→ m = v/u

→ m = -10/3/-5

→ m = + 2/3

The positive sign shows that the image is erect & virtual .

The image is Two - third of the size of object .

Answer 2

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Let's \; understand \; the \; concept \; 1^{st}}}}}}}$

This question says that an object is placed in front of a concave lens of focal length 10 cm at a distance of 5cm. Find the nature and position of image. Diagram needed.

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Given \; that}}}}}}$

An object is placed in front of a concave lens of focal length 10 cm at a distance of 5 cm. Means

Focal length = 10 cm

Distance = 5 cm

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{To \; find}}}}}}$

Nature and position of the image.

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Diagram}}}}}}$

In the attachment.

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Solution}}}}}}$

Nature and position of the image = erect and virtual

Position of the image is 3.33 cm

Nature of image is erect and virtual

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{They \; denotes}}}}}}$

u denotes distance of the object.

f denotes length of focal.

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Using \: formula}}}}}}$

$\: \: \: \: \: \:{\boxed{\bold{\bf{\longmapsto Len's \: formula \: = \: \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}}}$

$\: \: \: \: \: \:{\boxed{\bold{\bf{\longmapsto Magnification \: = \: \dfrac{v}{u}}}}}$

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Full \; Solution}}}}}}$

$\rule{150}{1}$

~ Finding the distance of the image

${\sf{\bold{\leadsto Len's \: formula \: = \: \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}}$

${\sf{\bold{\leadsto \dfrac{1}{v} = \dfrac{-1}{f} + \dfrac{1}{u}}}}$

${\sf{\bold{\leadsto \dfrac{1}{v} - \dfrac{-1}{10} = \dfrac{1}{-5}}}}$

${\sf{\bold{\leadsto \dfrac{1}{v} = -1-2/10}}}$

${\sf{\bold{\leadsto \dfrac{1}{v} = -3/10}}}$

${\sf{\bold{\leadsto v = \dfrac{-10}{3}}}}$

${\sf{\bold{\leadsto v = -33.3 cm}}}$

Image position is never negative henceforth,

${\sf{\bold{\leadsto v = 33.3 cm}}}$

${\pink{\frak{33.3 \: cm \: is \: position \: of \: image}}}$

$\rule{150}{1}$

~ Finding nature of the image,

${\sf{\bold{\leadsto m = \dfrac{v}{u}}}}$

${\leadsto}$ m = -10/3/-5

${\leadsto}$ m = ±2/3

${\leadsto}$ This mean the nature of image is erect and virtual

${\pink{\frak{Image \: is \: Erect \: and \: Virtual}}}$

$\: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Additional \: knowledge}}}}}}$

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.

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