Question

(1) If the magnification of a lens is +0,35. What will be the nature of the lens; Convex OR Concave. If the focal length of the lens lens is 15 cms; draw a ray diagram to study its image.​

Answer 1

Answer:

Focal length (f) = - 15cm

Distance of image (v) = -10cm

∴ 1/v - 1/u = 1/f

∴ 1/u = - (1/10) - (1/-15)

∴1/u = 1/15 - 1/10

∴ 1/u = -0.033

∴u = -30cm

So, the object is placed 30cm away from the concave lens.

Answer 2

Answer:

The nature of the lens is concave. The ray diagram is attached below.

Explanation:

Here we have been given to find the nature of the particular lens which produces a magnification of + 0.35

Now as we know a positive value of the magnification denotes the formation of a virtual and erect image. Here we can also see that the size of the resulting image has decreased by a factor of 0.35 which means that this image has been formed by the concave lens, As we know the concave lens always forms erect and virtual images which are smaller in size than that of the original object.

Here if the focal length of the particular concave lens is 15cm, we calculate the position of the image (v) and object (u) as below:

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

Here m is the magnification, v is the position of the image, and u is the position of the object taken.

∴ 0.35 = $\frac{v}{u}$

⇒ v = 0.35u (equation 1)

Now using the lens formula we calculate u and v as below:

Here f is the focal length, v is the position of the image, and u is the position of the object taken.

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

f = - 15cm (given)

v = 0.35u

substituting these in the above formula we get,

⇒ $\frac{1}{-15} =\frac{1}{0.35u} -\frac{1}{u}$

⇒  $\frac{1}{-15} =\frac{1-0.35}{0.35u}$

⇒ $\frac{1}{-15} =\frac{0.65}{0.35u}$

⇒ u = $\frac{0.65}{0.35}$ × (-15)

⇒ u = - 27.85 cm

Hence the position of the object is 27.85 cm in front of the lens.

Now v = 0.35 × u

∴ v = 0.35 × (-27.85) cm = - 9.75cm

Hence the image will be formed 9.75 cm in front of the lens.

Ray diagram for this is attached below. The position of the object is between the infinity and the respective optical center of the lens whereas the image lies between the focus and the optical center.