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ye tho out of syllabus question hai

Answer:

An object 1cm high produces a real image -1.5cm high when placed at a distance of 15cm from a concave miorror. The position of the image and the magnification will be 22.5 cm  and 1.5 cm.

GIVEN:

Height of object, ho = 1 cm

Height of the image, hi, is -1.5 cm.

Distance to Object, u = - 15 cm

I HAVE TO FIND:

either the image's position or its distance (v).

Magnification (m) (m).

Solution:

Finding the magnification is the solution.

Having said that,

[tex]\implies magnification, m=\frac{h_i}{h_o}[/tex]

By changing the values,

[tex]\implies magnification, \; m=\frac{-1.5}{1}[/tex]

[tex]\implies m=-1.5[/tex]

calculating the image's distance,

Having said that,

[tex]\implies magnification, \; m=-\frac{v}{u}[/tex]

By changing the values,

[tex]\implies -1.5=-\frac{v}{(-15)}[/tex]

[tex]\implies -1.5 \times -15=-v[/tex]

[tex]\implies -\frac{1.5}{10} \times -15=-v[/tex]

[tex]\implies -\frac{1.5}{2} \times -3=-v[/tex]

[tex]\implies -\frac{45}{2} =-v[/tex]

[tex]\implies 22.5=-v[/tex]

[tex]\implies v=-22.5\; cm[/tex]

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