Question

Ravi wants to get an inverted image of an object kept at a distance of 40cm from a concave mirror. The focal length of the mirror is 20cm.At what distance from the mirror should he place the screen?​

Answer 1

Answer:

• 40cm is the required distance from mirror to palace the screen .

Explanation:

$\rm {\underline{\pink{According~ to ~ the~ Question:-}}}$

It is given that,

• Focal length ,f = -20cm
• Object distance ,u = -40cm

We have to calculate image distance .

We use here the "Mirror Formula "

• $\boxed{\bf{\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}$

by putting the value we get

$\sf\implies\; \frac{1}{v} + \frac{1}{-40} = \frac{1}{-20} \\\\\sf\implies\; \frac{1}{v} - \frac{1}{40} = \frac{-1}{20} \\\\\sf\implies\; \frac{1}{v} = \frac{-1}{20} + \frac{1}{40} \\\\\sf\implies\; \frac{1}{v} = \frac{-2+1}{40} \\\\\sf\implies\; \frac{1}{v} = \frac{-1}{40} \\\\\sf\implies\; v = -40 cm \\\\\boxed{\bf{Hence,~the ~ image ~ distance ~ is ~ 40cm}}$

Extra Information !!

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

Answer 2

Answer:

Given :-

• Ravi wants to get an inverted image of an object kept at a distance of 40 cm from a concave mirror.
• The focal length of the mirror is 20 cm.

To Find :-

• What is the distance from the mirror should be placed at the screen.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\bigstar \sf\boxed{\bold{\pink{\dfrac{1}{v} + \dfrac{1}{u} =\: \dfrac{1}{f}}}}\: \: \: \bigstar$

where,

• v = Image Distance
• u = Object Distance
• f = Focal Length

Solution :-

Given :

• Focal Length (f) = - 20 cm
• Object Distance (u) = - 40 cm

According to the question by using the formula we get,

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

By putting those values we get,

$\implies \sf \dfrac{1}{v} + \bigg(- \dfrac{1}{40}\bigg) =\: \bigg(- \dfrac{1}{20}\bigg)$

$\implies \sf \dfrac{1}{v} - \dfrac{1}{40} =\: - \dfrac{1}{20}$

$\implies \sf \dfrac{1}{v} =\: - \dfrac{1}{20} + \dfrac{1}{40}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{- 2 + 1}{40}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{- 1}{40}$

By doing cross multiplication we get,

$\implies \sf v(- 1) =\: 40(1)$

$\implies \sf v \times (- 1) =\: 40 \times 1$

$\implies \sf - v =\: 40$

$\implies \sf\bold{\red{v =\: - 40\: cm}}$

$\sf\bold{\underline{\blue{\therefore\: The\: image\: distance\: is\: 40\: cm\: .}}}$