When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
Show Hint
- 40 cm away from the mirror.
- 80 cm away from the mirror.
- 20 cm towards the mirror.
- 20 cm away from the mirror.
The Correct Option is A
Approach Solution - 1
2. Magnification formula: \[ m = \frac{f}{f - u} \] \[ \frac{1}{2} = \frac{f}{f - (-40)} \] \[ f + 40 = 2f \implies f = 40 \mathrm{~cm} \]
3. For magnification $\frac{1}{3}$: \[ \frac{1}{3} = \frac{40}{40 - u} \] \[ 40 - u = 120 \implies u = -80 \mathrm{~cm} \] Therefore, the correct answer is (1) 40 cm away from the mirror.
Approach Solution -2
We are given that when an object is placed 40 cm away from a spherical mirror, an image with magnification \( m_1 = \frac{1}{2} \) is formed. We are asked to find how far the object must be moved to obtain an image with magnification \( m_2 = \frac{1}{3} \).
Concept Used:
For a spherical mirror, the mirror formula is:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
and the magnification \( m \) is given by:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \]
Thus, \( v = -mu \).
Step-by-Step Solution:
Step 1: Substitute \( v = -mu \) in the mirror formula.
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{u} + \frac{1}{-mu} \] \[ \frac{1}{f} = \frac{1 - \frac{1}{m}}{u} = \frac{m - 1}{mu} \] \[ f = \frac{mu}{m - 1} \]
Step 2: For the first condition, \( u_1 = 40 \, \mathrm{cm}, \, m_1 = \frac{1}{2} \).
\[ f = \frac{m_1 u_1}{m_1 - 1} = \frac{\frac{1}{2} \times 40}{\frac{1}{2} - 1} \] \[ f = \frac{20}{-\frac{1}{2}} = -40 \, \mathrm{cm} \]
Thus, the mirror has a focal length of \( -40 \, \mathrm{cm} \) (concave mirror).
Step 3: For the new magnification \( m_2 = \frac{1}{3} \), use the same formula to find \( u_2 \).
\[ f = \frac{m_2 u_2}{m_2 - 1} \] Substitute \( f = -40 \): \[ -40 = \frac{\frac{1}{3} u_2}{\frac{1}{3} - 1} = \frac{\frac{1}{3} u_2}{-\frac{2}{3}} = -\frac{u_2}{2} \] \[ u_2 = 80 \, \mathrm{cm} \]
Step 4: Compare the new and old object distances.
\[ u_1 = 40 \, \mathrm{cm}, \quad u_2 = 80 \, \mathrm{cm} \]
The object is moved farther from the mirror by:
\[ \Delta u = u_2 - u_1 = 80 - 40 = 40 \, \mathrm{cm} \]
Final Computation & Result:
Final Answer: The object should be moved 40 cm away from the mirror.
\[ \boxed{\text{Object is moved 40 cm away from the mirror.}} \]
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