Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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]

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.