Question

plzzzzzz help me.. ​

Answer 1

$\large{\red{\bold{\underline{Given:}}}}$

(i) f (Focal Length) = -5cm

(ii) u (Object Distance) = -10cm

(iii) h (Object height) = 6cm

$\large{\green{\bold{\underline{To \: Calculate:}}}}$

(i) v (image distance) = ????

(ii) h' (Size of the image) = ???

$\large{\blue{\bold{\underline{Formula \: Used:}}}}$

$\sf \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\large{\red{\bold{\underline{Calculation:}}}}$

$\rightarrow \sf \: \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \\ \\ \rightarrow \sf \: \frac{1}{v} = - \frac{1}{5} + ( - \frac{1}{10} ) \\ \\ \rightarrow \sf \: \frac{1}{v} = - \frac{1}{5} - \frac{1}{10} \\ \\ \rightarrow \sf \: \frac{1}{v} = \frac{ - 1 \times 2 - 1}{10} \\ \\ \rightarrow \sf \: \frac{1}{v} = \frac{ - 3}{10} \\ \\ \rightarrow \sf \: v = - 3.33cm$

$\large{\pink{\bold{\underline{Now:}}}}$

As per our information;-

$\sf \: \frac{Size \: of \: the \: Image}{Size \: of \: the \: object} = + \frac{v}{u}$

$\large{\green{\bold{\underline{On \: putting \: all \: the \: values:}}}}$

$\rightarrow \: \sf \: \frac{h'}{h} = \frac{ - 3.3}{ - 10} \\ \\ \rightarrow \: \sf \: \frac{h'}{6} = \frac{ \cancel- 3.3}{ \cancel- 10} \\ \\ \rightarrow \: \sf \: \frac{h'}{6} = \frac{3.3}{10} \\ \\ \rightarrow \: \sf \: h' = \frac{6 \times 3.3}{10} \\ \\ \rightarrow \: \sf \: h' = \frac{19.8}{10} \\ \\ \rightarrow \: \sf \: h' = 1.98$

$\large{\pink{\bold{\underline{Hence:}}}}$

Size of the image is 1.98cm.

$\large{\orange{\bold{\underline{Related \: Information:}}}}$

$\rightarrow \: \sf \: Always \: virtual \: image \: will \: be \:formed \: by \\ \sf \: concave \: lens. \\ \\ \rightarrow \: \sf \: Image \: will \: be \: erect. \\ \\ \rightarrow \: \sf \: Image \: will \: be \: formed \: always \: on \: the \: same \\ \sf \: side \: of \: the \: object.$

Answer 2

$\huge\mathcal{\mid{\mid{\underline{\green{Good\: Afternoon\:}}}{\mid{\mid}}}}$

Qᴜᴇsᴛɪᴏɴ ;-

❥︎ An object of height 6cm is placed perpendicular to the principle axis of a concave lens of focal length 5cm. Use the lens formula to determine the position, size and nature of the image if the distance from the lens is 10cm.

$\huge{\orange{\boxed{\fcolorbox{aqua}{indigo}{\color{lime}ANSWER}}}}$

$\Large\bf\pink{GiVeN,}$

• $\bf{\red{Height\:of\: object}}(h_i)\:=\:6\:cm\:$

• $\bf{\green{Focal\: length}}(f)\:=\:5\:cm\:$

• $\bf{\red{Object\: distance}}(u)\:=\:10\:cm\:$

$\Large\bf\orange{We\: know\: that,}$

$\blue\bigstar\:\boxed{\purple{\bf{\dfrac{1}{v}\:-\:\dfrac{1}{u}\:=\:\dfrac{1}{f}\:}}}$

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

$:\implies\:\sf{\dfrac{1}{v}\:=\:-\dfrac{1}{5}\:+\:\Big(\dfrac{1}{-10}\Big)\:}$

$:\implies\:\sf{\dfrac{1}{v}\:=\:-\dfrac{1}{5}\:-\:\dfrac{1}{10}\:}$

$:\implies\:\sf{\dfrac{1}{v}\:=\:\dfrac{-\:2\:-\:1}{10}\:}$

$:\implies\:\sf{\dfrac{1}{v}\:=\:\dfrac{-3}{10}\:}$

$:\implies\:\sf{v\:=\:\dfrac{-10}{3}}$

$:\implies\:\bf\green{v\:=\:-3.33\:cm}$

$\Large\bf\red{Also\:we\: have,}$

$\orange\bigstar\:\boxed{\purple{\bf{Magnification\:=\:\dfrac{v}{u}\:=\:\dfrac{h_i}{h_o}\:}}}$

$:\implies\:\sf{\dfrac{-\frac{10}{3}}{-10}\:=\:\dfrac{h_i}{6}\:}$

$:\implies\:\sf{\dfrac{1}{3}\times{6}\:=\:h_i\:}$

$:\implies\:\bf{h_i\:=\:{2}\:cm}$

$\Large\bold\therefore$ The image of size 2cm will be formed infront of the lens at a distance of 3.33cm from the lens. Thus, the nature of image is virtual and erect.