Question

$\huge \mathfrak \color{Purple}{\underline{\underline{Question }}} :$
A concave lens of focal length 15 cm forms an image 10cm from the lens. How far is the object placed from the lens ? Draw the ray diagram .
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Answer 1

$\huge \sf \color{Purple}{\underline{\underline{\underline{\dag Answer} }}} :$

The object is placed at 30 cm from the lens and For ray diagram refer to attachment .

$\huge \sf \color{Purple}{\underline{\underline{\underline{\dag Given} }}} :$

☆Focal length of concave lens, f = -15 cm

☆distance of the image, v = -10 cm

$\boxed{\sf Sign \ of \ v \ has \ taken \ -ve \ because \ a \ concave \ lens \ always \ forms \ a \ virtual \ image }$

$\huge \sf \color{Purple}{\underline{\underline{\underline{\dag To \ find } }}} :$

Distance between lens and the object

$\huge \sf \color{Purple}{\underline{\underline{\underline{\dag Solution} }}} :$

$\large \frac{1}{v}-\frac{1}{u} = \frac{1}{f} \\\\ \frac{1}{u} = \frac{1}{v} -\frac{1}{f} \\ \\ \implies \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} \\ \\ \implies \frac{1}{u} = \frac{1}{-10} + \frac{1}{15} \\ \\ \implies \frac{1}{u} = \frac{-3+2}{30} \\ \\ \implies \frac{1}{u} = \frac{1}{30} \\ \\ \boxed{U = -30 cm }$

Thus, the object is placed at 30 cm from the lens . The ray diagram provided in attachment .