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An object is placed at a distance of 30 cm from a concave lens of focal length 30 cm.

(i) Use lens formula to determine the distance of the image from the lens.

(ii) List four characteristics of the image (nature position, size, erect/inverted) in this case.

(iii) Draw a labelled diagram to justify your answer of part (ii).

 

 

 

 

 
 
 
 
 

Answers (1)

u=-30 \; cm

f=-30 \; cm

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

\frac{1}{-30}=\frac{1}{v}-(\frac{1}{-30})

\frac{1}{-30}-\frac{1}{30}=\frac{1}{v}

\frac{1-1}{30}=\frac{1}{v}

v=-15 \; cm

(ii) Nature \rightarrowVirtual

Position \rightarrow 15 cms from lens on the same side of the object.

Errect/Inverted \rightarrowErrect

m=\frac{v}{u}=\frac{-15}{-30}=\frac{1}{2}

Size \rightarrow Diminished

Posted by

Sumit Saini

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