Respuesta :
The nature of the image is magnified , virtual , and upright.
Further explanation
We will solve this problem using following formula:
[tex]\large{ \boxed {\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}}}[/tex]
where:
so = distance of object from mirror
si = distance of image from mirror
f = focal length of mirror
[tex]\texttt{ }[/tex]
Given:
Magnification = M = 1
Initial Image Distance = si = 30 cm
Difference of Object's Distance = Δso = 20 cm
Unknown:
Final Image Distance = si' = ?
Solution:
Initial Condition:
[tex]M = \frac{s_i}{s_o}[/tex]
[tex]1 = \frac{s_i}{s_o}[/tex]
[tex]s_i = s_o[/tex]
[tex]s_o = 30 cm[/tex]
[tex]\texttt{ }[/tex]
[tex]\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}[/tex]
[tex]\frac{1}{s_o} + \frac{1}{s_o} = \frac{1}{f}[/tex]
[tex]\frac{2}{s_o} = \frac{1}{f}[/tex]
[tex]f = \frac{1}{2} s_o[/tex]
[tex]f = \frac{1}{2} \times 30[/tex]
[tex]f = 15 \texttt{ cm}[/tex]
[tex]\texttt{ }[/tex]
Final Condition:
[tex]s_o' = s_o - \Delta s[/tex]
[tex]s_o' = 30 - 20[/tex]
[tex]s_o' = 10 \texttt{ cm}[/tex]
[tex]\texttt{ }[/tex]
[tex]\frac{1}{s_o'} + \frac{1}{s_i'} = \frac{1}{f}[/tex]
[tex]\frac{1}{10} + \frac{1}{s_i'} = \frac{1}{15}[/tex]
[tex]\frac{1}{s_i'} = \frac{1}{15} - \frac{1}{10}[/tex]
[tex]\frac{1}{s_i'} = \frac{2}{30} - \frac{3}{30}[/tex]
[tex]\frac{1}{s_i'} = -\frac{1}{30}[/tex]
[tex]s_i' = 30 \texttt{ cm}[/tex]
Because si' < 0 , the nature of the image will be magnified , virtual , and upright.
[tex]\texttt{ }[/tex]
Learn more
- Compound Microscope : https://brainly.com/question/7512369
- Reflecting Telescope : https://brainly.com/question/12583524
- Focal Length : https://brainly.com/question/8679241
- Mirror and Lenses : https://brainly.com/question/3067085
Answer details
Grade: High School
Subject: Physics
Chapter: Light
Keywords: Light , Mirror , Magnification , Ray , Diagram , Image , Real , Virtual
The final image formed after shifting the object will be virtualand erect image and formed at a distance of [tex]\boxed{30\,{\text{cm}}}[/tex] behind the mirror.
Further Explanation:
Given:
The magnification of the image for the object placed at [tex]30\text{cm}[/tex] from the mirror is [tex]-1[/tex].
The final distance of the object from the mirror is [tex]30\,{\text{cm}} - 20\,{\text{cm}} = 10\,{\text{cm}}[/tex] .
Concept:
The magnification of the image for the object placed at [tex]30\text{cm}[/tex] from the mirror is [tex]-1[/tex].
The magnification of the image can be represented as.
[tex]m = - \dfrac{{{d_i}}}{{{d_o}}}[/tex]
Substitute [tex]-1[/tex] for [tex]m[/tex] and [tex]-30\text{cm}[/tex] for [tex]{d_o}[/tex] in above expression.
[tex]\begin{aligned}- 1 &= - \frac{{{d_i}}}{{ - 30}}\\{d_i} &= - 30\,{\text{cm}}\\\end{aligned}[/tex]
Here, the image distance is equal to the object distance. In case of the mirror, the image distance and the object distance are equal only when the object is placed at the centre of curvature of the mirror.
Therefore, the focal length of the mirror will be half of the distance of the object from the mirror.
[tex]\begin{aligned}f&= \frac{{{d_o}}}{2}\\&= \frac{{ - 30}}{2}\,{\text{cm}}\\&= - {\text{15}}\,{\text{cm}}\\\end{aligned}[/tex]
Now, the object is shifted towards the mirror. The new distance of the object from the mirror is [tex]10\text{cm}[/tex].
The mirror’s formula can be expressed as.
[tex]\dfrac{1}{{{d_i}}} + \dfrac{1}{{{d_o}}} = \dfrac{1}{f}[/tex]
Substitute [tex]-10\text{cm}[/tex] for [tex]d_o[/tex] and [tex]-15\text{cm}[/tex] for [tex]f[/tex] in above equation.
[tex]\begin{aligned}\frac{1}{{{d_i}}} + \frac{1}{{ - 10}} &= \frac{1}{{ - 15}}\\\frac{1}{{{d_i}}} &= \frac{1}{{10}} - \frac{1}{{15}}\\{d_i} &= 30\,{\text{cm}}\\\end{aligned}[/tex]
The image distance is positive. It means that the final image of the object will be formed behind the mirror i.e. the image will be virtual.
The magnification of the image is.
[tex]\begin{aligned}m&= - \frac{{{d_i}}}{{{d_o}}}\\&= - \frac{{30}}{{ - 10}}\\ &= + 3\\\end{aligned}[/tex]
Thus, the image formed will be erect, virtual and three times the size of the object.
Learn More:
- What type of mirror do dentist use https://brainly.com/question/997618
- A small bulb has a resistance of 2 ohm when cold https://brainly.com/question/10421964
- The heavier elements in the universe were formed by https://brainly.com/question/4581837
Answer Details:
Grade: Middle School
Chapter: Mirror’s Formula
Subject: Physics
Keywords: Image distance, object distance, mirror’s formula, magnification, 30 cm from the mirror, now moved 20cm, ray diagram, real inverted.
