Question

An object is placed 15 cm from a diverging mirror that has a focal length of 10 cm. What will be the image distance.​

Answer 1

Answer:

• $\sf{Distance\:of\:the\: image\:is\:6\: cm}$

Explanation:

${\underline{\underline{\sf{\red{{\bigstar}\:\:\:Given:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\green{ An\: object \:is\: placed \:15 cm\: from \:a\: diverging\: mirror\: that\: has\: a \:focal\: length \:of\: 10 cm.}$

${\underline{\underline{\sf{\orange{{\bigstar}\:\:\:ToFind:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\purple{Image \ distance}$

${\underline{\underline{\sf{\blue{{\bigstar}\:\:\: Solution:-}}}}}$

$\tiny\red{\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} =\: \dfrac{1}{f}}}}$

$\tiny{\underline{\sf{\:\:\:\:Where,\:\:\:}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\orange{u = Object \ Distance}\\\tiny\:\:\:\:\bullet\:\:\:\sf\green{v = Image \ Distance}\\\tiny\:\:\:\:\bullet\:\:\:\sf\blue{f = Focal \ Length}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ \dfrac{1}{v} =\: \dfrac{1}{f} - \dfrac{1}{u}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{\dfrac{1}{v} =\: \dfrac{1}{10} - (- \dfrac{1}{15}) }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{\dfrac{1}{v} =\: \dfrac{1}{10} + \dfrac{1}{15} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{\dfrac{1}{v} =\: \dfrac{3 + 2}{30} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{\dfrac{1}{v} =\: \dfrac{5}{30} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ \dfrac{1}{v} =\: \dfrac{\cancel{5}}{\cancel{30}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ \dfrac{1}{v} =\: \dfrac{1}{6}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{v =\: 1 \times 6 }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ v =\: 6\: cm}$