Question

and object is placed at 30 cm before a concave mirror of focal length of 20cm to get a real image what will be the distance of the image from the mirror​

Answer 1

$\huge \purple{ \mathfrak{solution}}$

Given:

• object distance (u) = -30cm

• Focal length(f)= -20cm

To find :

• Image distance (v)= ?

By using mirror formula :

$\boxed{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

Now put the given values :

$\rightarrow \: \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ \rightarrow \frac{1}{v} = \frac{ - 1}{20} + \frac{1}{30} \\ \\ \rightarrow \: \frac{1}{v} = \frac{ - 3 + 2}{60} \\ \\ \rightarrow \: \frac{1}{v} = - \frac{1}{60} \\ \\ \rightarrow \: v \: = - 60$

Hence the image distance is

$\huge \green{\boxed{ \mathfrak{ - 60cm}}}$

Answer 2

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

The image will be formed between c and ∞ , real and inverted , size of image is larger than object

Given :-

→object distance (u) = -30cm

→Focal length(f)= -20cm

To find :-

→ Image distance (v)= ?

So, now we can jse mirror formula to find the image distance .

$\huge \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

So, now we putting thw values in formula .

$\sf{ \implies \: \frac{1}{v} = \frac{1}{f} - \frac{1}{u} } \\ \\ \sf{ \implies \: \frac{1}{v} = \frac { - 1}{20} + \frac{1}{30} } \\ \\ \sf{ \implies \: \frac{1}{v} = \frac{ - 3 + 2}{ 60} } \\ \\ \sf{ \implies \: \frac{1}{v} = \frac{ - 1}{ 60} } \\ \\ \sf{ \implies \: v \: = - 60 \: }$

Hence , the image distance is -60