Question

the distance between an object and a concave lens is 18 cm . Focal length is 6cm . Find nature, position and magnification of image using lens formula
​

Answer 1

Answer:

By lens formula:

v

1

​

−  

u

1

​

=  

f

1

​

 

[4pt]

⇒  

−15 cm

1

​

−  

u

1

​

=  

−20 cm

1

​

 

[4pt]

⇒  

u

1

​

=  

20 cm

1

​

+  

−15 cm

1

​

 

[4pt]

⇒  

u

1

​

=  

60 cm

3−4

​

=  

60 cm

−1

​

 

[4pt]

⇒u=−60 cm

Therefore, object is placed at 60 cm away from the lens, on the same side as image.

Now,

Height of object, h  

1

​

=5cm

Magnification, m=  

h  

1

​

 

h  

2

​

 

​

=  

u

v

​

 

Putting value of v and u:

Magnification, m=  

5 cm

h  

2

​

 

​

=  

−60 cm

−15 cm

​

 

⇒  

5 cm

h  

2

​

 

​

=  

4

1

​

 

⇒h  

2

​

=  

4

5 cm

​

=1.25 cm

Thus, the height of the image is 1.25 cm and the positive sign means the image is virtual and erect.

Explanation:

Hope it helped

Answer 2

The mirror formula is 1+1=1,
1
u
+
1
v
=
1
f
,
where,


u
is the distance of the object from the mirror,


v
is the distance of the image from the mirror, and,


f
is the focal length of the mirror.

As per the sign convention, distances to the right of the mirror are termed positive and the distances to the left are termed negative, with the reflecting surface of the mirror facing the left.

In concave mirrors the the object, image and focus are all to the left of the mirror. Hence the values of all these distances have a negative sign.

It is given that =−6
u
=
−
6
cm and =−18
v
=
−
18
cm.

⇒−16−118=1⇒1=−29.
⇒
−
1
6
−
1
18
=
1
f
⇒
1
f
=
−
2
9
.

⇒=−92=−4.5
⇒
f
=
−
9
2
=
−
4.5
cm.

⇒
⇒
The focal length is 4.5
4.5
cm.