Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $\left(\frac{\mathrm{x}}{4}\right) \mathrm{cm}$, where magnitude of value of x is _______ .
Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $\left(\frac{\mathrm{x}}{4}\right) \mathrm{cm}$, where magnitude of value of x is _______ .
Show Hint
Correct Answer: 45
Approach Solution - 1
2. Distance between object and image: \[ u - v = 30 \mathrm{~cm} \] \[ u - \frac{u}{3} = 30 \implies u = 45 \mathrm{~cm}, \quad v = 15 \mathrm{~cm} \]
3. Focal length: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} + \frac{1}{45} = \frac{4}{60} \] \[ f = \frac{60}{4} = 15 \mathrm{~cm} \] \[ x = 45 \] Therefore, the correct answer is (45).
Approach Solution -2
We are given that the magnification produced by a mirror is \( m = -\frac{1}{3} \) and the distance between the object and its image is \( 30 \, \mathrm{cm} \). We need to find the focal length \( f \) of the mirror, expressed as \( \frac{x}{4} \, \mathrm{cm} \), and determine the magnitude of \( x \).
Concept Used:
For a mirror, the magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
and the mirror formula is:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
The distance between object and image is:
\[ |v - u| = 30 \, \mathrm{cm} \]
Step-by-Step Solution:
Step 1: From magnification \( m = -\frac{v}{u} \), substitute \( m = -\frac{1}{3} \):
\[ -\frac{1}{3} = -\frac{v}{u} \] \[ \Rightarrow v = \frac{u}{3} \]
Step 2: The distance between the object and image is 30 cm.
\[ |v - u| = 30 \]
Substitute \( v = \frac{u}{3} \):
\[ \left| \frac{u}{3} - u \right| = 30 \] \[ \left| \frac{u - 3u}{3} \right| = 30 \] \[ \frac{2|u|}{3} = 30 \]
Step 3: Solve for \( u \):
\[ |u| = 45 \, \mathrm{cm} \]
Since the object is in front of the mirror, \( u = -45 \, \mathrm{cm} \).
Step 4: Substitute in \( v = \frac{u}{3} \):
\[ v = \frac{-45}{3} = -15 \, \mathrm{cm} \]
Step 5: Apply the mirror formula:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] \[ \frac{1}{f} = \frac{1}{-45} + \frac{1}{-15} = -\left(\frac{1}{45} + \frac{3}{45}\right) = -\frac{4}{45} \] \[ f = -\frac{45}{4} \, \mathrm{cm} \]
Step 6:
The focal length is given as \( \frac{x}{4} \, \mathrm{cm} \), so comparing:
\[ \frac{x}{4} = -\frac{45}{4} \Rightarrow x = -45 \]
Final Computation & Result:
The magnitude of \( x \) is:
\[ \boxed{|x| = 45} \]
Final Answer: \( x = 45 \)
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