Focal length of a convex mirror is 15 cm what will be the maximum distance of image formed by mirror

Concept:

Focal length (f):

Focal length is the distance between the center of a convex lens or a concave mirror and the focal point of the lens or mirror i.e. the point where parallel rays of light meet.

Using the formula

\(\frac{1}{{\rm{v}}} - \frac{1}{{\rm{u}}} = \frac{1}{{\rm{f}}}\) the velocity v can be found as u and v are given above.

Calculation:

We know that the focal length would be one-half the radius of curvature

i.e. R  = 2f

20 cm = 2f

f = 10 cm

The arrangement of convex lens and concave mirror is:

The lens formula is:

\(\frac{1}{{\rm{v}}} - \frac{1}{{\rm{u}}} = \frac{1}{{\rm{f}}}\)

Where, v = Distance of image from optical centre of lens

Given, Distance of object from optical centre of lens, u = -30 cm

Focal length of lens, f = 20 cm

Now, substituting the values,

\( \Rightarrow \frac{1}{{20}} = \frac{1}{{\rm{v}}} - \frac{1}{{\left( { - 30} \right)}}\)

\( \Rightarrow \frac{1}{{20}} - \frac{1}{{30}} = \frac{1}{{\rm{v}}}\)

\( \Rightarrow \frac{1}{{\rm{v}}} = \frac{{30 - 20}}{{600}}\)

\( \Rightarrow \frac{1}{{\rm{v}}} = \frac{{10}}{{600}}\)

\( \Rightarrow \frac{1}{{\rm{v}}} = \frac{1}{{60}}\)

∴ v = 60 cm

A virtual image is formed when an object is located less than one focal length from a concave mirror (i.e., in front of focal point).

Since, we know that, the focal point is the midpoint of the line segment adjoining the vertex and the centre of curvature, the focal length would be one-half the radius of curvature.

R = 2f

According to the condition, image formed by lens should be the centre of curvature (radius of the curvature) of the mirror.

So, 20 cm = 2f

⇒ f = 10 cm

The distance from the mirror to the focal point is known as the focal length.

Thus, in front of focal point is 10 cm which is maximum distance of the object for which this concave mirror, by itself would produce a virtual image.

A concave lens has focal length of 15 cm. At what, distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification of the lens.

A concave lens always forms a virtual, erect image on the same side of the object. Given, Image distance, v = -10 cmFocal length,    f = - 15 cm            [f is -ve for a concave lens]Object distance, u = ? Now, using lens formula, 

                        1v-1u =1f

                       1u =1v-1f        = 1-10 -1-15        = -3+230       = -130 

i.e.,                    u = -30 cm 

Magnification, m= vu = -10-30 = +13 = + 0.33