When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is -1/2. Where should the object be placed to get a magnification of -1/5?

To solve this problem, we will use the mirror equation and the magnification formula.
Mirror Equation:
The mirror equation is given by:
1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
Where:
• ff is the focal length of the concave mirror,
• uu is the object distance (distance of the object from the mirror),
• vv is the image distance (distance of the image from the mirror).
Magnification Formula:
The magnification mm is given by:
m=Image heightObject height=vum = \frac{\text{Image height}}{\text{Object height}} = \frac{v}{u}
Where:
• mm is the magnification,
• vv is the image distance,
• uu is the object distance.
Step 1: Use the initial condition to find the focal length
We are given that when the object is placed at a distance of u1=50 cmu_1 = 50 \, \text{cm}, the magnification produced is m1=−12m_1 = -\frac{1}{2}.
Using the magnification formula:
m1=v1u1=−12m_1 = \frac{v_1}{u_1} = -\frac{1}{2}
Thus, the image distance v1v_1 can be found:
v1=m1×u1=−12×50=−25 cmv_1 = m_1 \times u_1 = -\frac{1}{2} \times 50 = -25 \, \text{cm}
Now, we can use the mirror equation to find the focal length ff:
1f=1u1+1v1\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1}
Substitute the values:
1f=150+1−25\frac{1}{f} = \frac{1}{50} + \frac{1}{-25} 1f=150−125\frac{1}{f} = \frac{1}{50} - \frac{1}{25} 1f=150−250=−150\frac{1}{f} = \frac{1}{50} - \frac{2}{50} = -\frac{1}{50}
Thus, the focal length ff is:
f=−50 cmf = -50 \, \text{cm}
Step 2: Find the new object distance for a magnification of −15-\frac{1}{5}
Now, we are asked to find the object distance u2u_2 when the magnification m2=−15m_2 = -\frac{1}{5}.
Using the magnification formula:
m2=v2u2=−15m_2 = \frac{v_2}{u_2} = -\frac{1}{5}
Thus, the image distance v2v_2 can be found:
v2=m2×u2=−15×u2v_2 = m_2 \times u_2 = -\frac{1}{5} \times u_2
Now, use the mirror equation again:
1f=1u2+1v2\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2}
Substitute v2=−15u2v_2 = -\frac{1}{5} u_2 and f=−50f = -50:
1−50=1u2+1−15u2\frac{1}{-50} = \frac{1}{u_2} + \frac{1}{-\frac{1}{5} u_2}
Simplify the second term:
1−50=1u2−5u2\frac{1}{-50} = \frac{1}{u_2} - \frac{5}{u_2} 1−50=−4u2\frac{1}{-50} = \frac{-4}{u_2}
Now, solve for u2u_2:
u2=−4×50=−200 cmu_2 = -4 \times 50 = -200 \, \text{cm}
Final Answer:
To get a magnification of −15-\frac{1}{5}, the object should be placed at a distance of 200 cm from the concave mirror.