If x + 1/x =4 , find
![{x}^{2} + \frac{1}{ {x}^{2} } {x}^{2} + \frac{1}{ {x}^{2} }](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B2%7D+++%2B+++%5Cfrac%7B1%7D%7B+%7Bx%7D%5E%7B2%7D+%7D+)
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Answer:
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Answer:
Solution
\Large{\underline{\mathfrak{\bf{Given}}}}
Given
Distance of object from concave mirror (U) = -10 cm
Radius of curvature of concave mirror ( R ) = -8 cm = 2F
So, F = -8/2 = -4 cm
\Large{\underline{\mathfrak{\bf{Find}}}}
Find
Distance of image (V) .
\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}
Explanation
\Large{\underline{\mathfrak{\bf{Using\:mirror\:formula}}}}
Usingmirrorformula
\red{\boxed{\sf{\orange{\:\dfrac{1}{F}\:=\:\dfrac{1}{U}\:+\:\dfrac{1}{V}}}}}
F
1
=
U
1
+
V
1
Substitute value of U and F in above equation ,
\begin{gathered}:\implies\sf{\:\dfrac{1}{V}\:+\:\dfrac{1}{-10}\:=\:\dfrac{1}{-4}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:-\:\dfrac{1}{4}\:+\:\dfrac{1}{10}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-10+4}{40}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-14}{40}} \\ \\ :\implies\sf{\:V\:=\:\dfrac{-40}{14}} \\ \\ :\implies\sf{\red{\:V\:=\:\dfrac{-20}{7}\:cm}}\end{gathered}
:⟹
V
1
+
−10
1
=
−4
1
:⟹
V
1
=−
4
1
+
10
1
:⟹
V
1
=
40
−10+4
:⟹
V
1
=
40
−14
:⟹V=
14
−40
:⟹V=
7
−20
cm
\Large{\underline{\mathfrak{\bf{Thus}}}}
Thus
Distance of image (V ) = -20/7 cm
When, the image distance is negative , the image is behind the mirror .
So, the image is virtual and upright.