(a) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens.(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculatei. The distance of the object from the lens.ii. The magnification for the image formediii. The nature of the image formed.

(a) Ray diagram showing the formation of image of an object placed between infinity and the optical centre $(C)$ of a concave lens.

When an object is placed between infinity and the optical centre $(C)$ of a concave lens, then the image formed is virtual, erect, highly diminished, and in front of the lens.

(b) Given:

Focal length, $f$ = $-$15 cm   (focal length of a concave lens is always taken negative)

Image distance, $v$ = $-$10 cm      (as the image is formed on the left side of the lens, it is taken negative)

To find: (i) The distance of the object from the lens, $u$.

              (ii) The magnification for the image formed, $m$.

             (iii) The nature of the image formed.

Solution: 

(i) From the lens formula we know that-

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

Substituting the given values we get-

$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-15)}$

$-\frac {1}{10}-\frac {1}{u}=-frac {1}{15}$

$\frac {1}{u}=\frac {1}{15}-\frac {1}{10}$

$\frac {1}{u}=\frac {2-3}{30}$

$\frac {1}{u}=-\frac {1}{30}$

$u=-30cm$

Thus, the object, $u$ is at a distance of 30cm from the concave lens (on the left side).

(ii) From the magnification formula we know that-

$m=\frac {v}{u}$

Substituting the value of $v$ and $u$ we get-

$m=\frac {-10}{-30}$

$m=\frac {1}{3}$

$m=+0.33$

Thus, the magnification, $m$ for the image formed is +0.33.

(iii) The nature of the image formed is virtual, erect, and diminished. Since the value of magnification is less than 1 (it is 0.33), therefore the image is smaller than the object  (or diminished). The plus sign $(+)$ for the magnification implies that the image is virtual and erect.

Also, we know that a concave lens always forms virtual, erect, and diminished images.

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Updated on: 10-Oct-2022

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