ANSWER THE FOLLOWING QUESTION:-
At what distance from a concave mirror of focal length 85 cm should an object 3 cm long be placed in order to get an erect image 7cm tall ?
Answers
Answer:
The object should be placed at a distance of 10 cm from the concave mirror .
Explanation :-
Firstly, according to the New cartesian sign convention (for a concave mirror), we have :-
→ f = - 15 cm
→ hₒ = 2 cm
→ hᵢ = 6 cm
Now, we will calcualte the relationship between the object and image distance by putting the values in formula of magnification of a spherical mirror.
\boxed{\bf{ \textbf{m} = \dfrac{-v}{u} = \dfrac{h_i}{h_o}}}
m=
u
−v
=
h
o
h
i
\begin{gathered} \implies \dfrac{-v}{u} = \dfrac{6}{2}\\\\\:\implies \dfrac{-v}{u} = 3\\\\\implies -v = 3u\\\\\:\implies v = -3u\end{gathered}
⟹
u
−v
=
2
6
⟹
u
−v
=3
⟹−v=3u
⟹v=−3u
Finally, let's calculate the required position of the object by putting values in the mirror formula.
\begin{gathered}\underline{ \boxed{\bf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}} \\\end{gathered}
v
1
+
u
1
=
f
1
\begin{gathered} \implies \dfrac{1}{-3u} + \dfrac{1}{u} = \dfrac{1}{-15}\\\\\:\implies \dfrac{-1}{3u} + \dfrac{1}{u} = \dfrac{-1}{15}\\\\\:\implies \dfrac{-1 + 3}{3u} = \dfrac{-1}{15}\\\\\:\implies \dfrac {2}{3u} = \dfrac{-1}{15}\\\\\:\implies -3u = 30\\\\\:\implies u = \dfrac{30}{-3}\\\\\:\implies \underline{u = -10 \: \: cm}\end{gathered}
⟹
−3u
1
+
u
1
=
−15
1
⟹
3u
−1
+
u
1
=
15
−1
⟹
3u
−1+3
=
15
−1
⟹
3u
2
=
15
−1
⟹−3u=30
⟹u=
−3
30
⟹
u=−10cm
Explanation:
Hii
can I get your intro
Explanation:
Focal length of concave mirror (f) = 85 cm
Height of the object (ho) = 3cm
Height of the image (hi) = 7cm
We need to find the object distance
So we know that
Magnification m = hi/ho
hi/ho = 7/3
hi/ho = 2.33 (erect image)
m = -v/u
2.33 = -v/u
⇒ v = -2.33u
Let us now consider Mirror formula
1/v + 1/u = 1/f
1/(-2.33u) + 1/u = 1/(85)
⇒ u = -13/23*85
u = 48.04 cm
distance of concave mirror is 48.04cm