If the image formed by a lens for all positions of an object placed in front it is always erect and diminished What is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power of this lens of 10 D, What is its focal length in the Cartesian system ?

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Solution

We know,
$P (P o w e r) = \frac{1}{f (f o c a l l e n g t h)}$
$\Rightarrow 10 = \frac{1}{f}$
$\Rightarrow f = 0.1 = 10 c m$
As the image formed is always erect and dimished, lens is concave lens. So,
$f = - 10 c m$

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If the image formed by a lens for all positions of an object placed in front it is always erect and diminished What is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power of this lens of 10 D, What is its focal length in the Cartesian system ?

We know,P(Power)=1f(focal length)⇒10=1f⇒f=0.1=10cmAs the image formed is always erect and dimished, lens is concave lens. So,f=−10cm

We know,
$P (P o w e r) = \frac{1}{f (f o c a l l e n g t h)}$
$\Rightarrow 10 = \frac{1}{f}$
$\Rightarrow f = 0.1 = 10 c m$
As the image formed is always erect and dimished, lens is concave lens. So,
$f = - 10 c m$