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a diverging lens has a focal length of -14 cm. An object is placed 38 cm from the lens's surface. determine the image distance

Sagot :

  • Focal length=-14cm=f
  • Object distance=u=38cm
  • Image distance=v

Using lens makers formula

[tex]\\ \tt\longmapsto \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

[tex]\\ \tt\longmapsto \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]

[tex]\\ \tt\longmapsto \dfrac{1}{v}=\dfrac{1}{-14}-\dfrac{1}{38}[/tex]

[tex]\\ \tt\longmapsto \dfrac{1}{v}=\dfrac{-19-7}{266}[/tex]

[tex]\\ \tt\longmapsto \dfrac{1}{v}=\dfrac{-26}{266}[/tex]

[tex]\\ \tt\longmapsto v=\dfrac{266}{-26}[/tex]

[tex]\\ \tt\longmapsto v=\dfrac{133}{-13}[/tex]

[tex]\\ \tt\longmapsto v=-10.2cm[/tex]

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