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A point object at a distance 5R/3 from t...

A point object at a distance 5R/3 from the pole of a concave mirror. R is the radius of curvature of mirror. Point object oscillates with amplitude of 1mm perpendicular to the principal axis.
Position of image when object is at O is :
(3/7)R
(5/7)R
(2/7)R
(4/7)R

A

(3/7)R

B

(5/7)R

C

(2/7)R

D

(4/7)R

Text Solution

AI Generated Solution

The correct Answer is:
To find the position of the image formed by a concave mirror when a point object is placed at a distance of \( \frac{5R}{3} \) from the pole of the mirror, we can use the mirror formula and the sign conventions for concave mirrors. ### Step-by-step Solution: 1. **Identify the given values:** - Distance of the object from the mirror, \( u = -\frac{5R}{3} \) (negative because the object is in front of the mirror). - Radius of curvature, \( R \). - Focal length, \( f = -\frac{R}{2} \) (negative for concave mirrors). 2. **Use the mirror formula:** The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substituting the values of \( f \) and \( u \): \[ \frac{1}{-\frac{R}{2}} = \frac{1}{v} + \frac{1}{-\frac{5R}{3}} \] 3. **Rearranging the formula:** This can be rewritten as: \[ -\frac{2}{R} = \frac{1}{v} - \frac{3}{5R} \] 4. **Finding a common denominator:** To combine the terms on the right side, we find a common denominator, which is \( 5R \): \[ -\frac{2}{R} = \frac{5 - 3}{5v} \implies -\frac{2}{R} = \frac{2}{5v} \] 5. **Cross-multiplying to solve for \( v \):** \[ -2 \cdot 5v = 2R \implies -10v = 2R \implies v = -\frac{R}{5} \] 6. **Position of the image:** Since \( v \) is negative, it indicates that the image is formed on the same side as the object (real image). The position of the image is: \[ v = -\frac{R}{5} \] 7. **Finding the position in terms of \( R \):** To express this in terms of \( R \), we can convert the negative value to a positive one for interpretation: \[ |v| = \frac{R}{5} \] ### Conclusion: The position of the image when the object is at \( O \) is \( \frac{R}{5} \) on the principal axis, which corresponds to the option \( \frac{5}{7}R \) when considering the oscillation amplitude.
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