a diverging mirror of 52.0 focal lenght produces a virtual image of 26.0cm from the mirror.
how far from the mirror should the object be placed?
solution
f=-52.0cm, v=-26.0cm
1/u + 1/v = 1/f
rearranging the terms
1/u = 1/f - 1/v
substituting the values of v and f into the above eq. we get
1/u = 1/-52.0 - 1/-26.0
ans = 52cm
am i right pls check.
Since it is a diverging mirror, the focal length is -52 cm, Since the image is virtual, the imaage distance is -26 cm
Do the calculation again and solve for u
To solve for the distance from the mirror (object distance, u), you correctly applied the mirror formula:
1/u + 1/v = 1/f
By substituting the values for f and v, you obtained:
1/u = 1/-52.0 - 1/-26.0
Now, let's simplify this equation. To add fractions, we need a common denominator:
1/u = (-1/-52.0) + (-1/-26.0)
Simplifying further:
1/u = 1/52.0 + 1/26.0
To add fractions with different denominators, we need to find a common denominator. In this case, the common denominator is 52:
1/u = 2/52.0 + 4/52.0
Adding the fractions together:
1/u = 6/52.0
Now, invert both sides of the equation:
u/1 = 52.0/6
Simplifying:
u = 52.0/6
Evaluating this division:
u ≈ 8.67 cm
Therefore, the object should be placed approximately 8.67 cm from the mirror.